Categories of Disabilities and Their Ambiguities
So far I have said a lot about why inclusion has come to be important for teachers, but not much about the actual nature of students' disabilities. Part of the reason for delaying was because, to put it simply, disabilities are inherently ambiguous. Naming and describing "types" of them implies that disabilities are relatively fixed, stable, and distinct, like different kinds of fruit or vegetables. As many teachers discover, though, the reality is somewhat different. The behavior and qualities of a particular student with a disability can be hard to categorize. The student may be challenged not only by the disability, but also by experiences common to all students, disabled or not. Any particular disability, furthermore, poses problems more in some situations than in others. A student with a reading difficulty may have trouble in a language arts class, for example, but not in a physical education class; a student with a hearing impairment may have more trouble "hearing" a topic that he dislikes compared to one that he likes. Because official descriptions of types or categories of disabilities overlook these complexities, they risk stereotyping the real, live people to whom they are applied (Green, et al., 2005). Even the simplifications might not be a serious problem if the resulting stereotypes were complimentary— most of us would not mind being called a "genius", for example, even if the description is not always true. Stereotypes about disabilities, however, are usually stigmatizing, not complimentary.
Still, categories of disabilities do serve useful purposes by giving teachers, parents, and other professionals a language or frame of reference for talking about disabilities. They also can help educators when arranging special support services for students, since a student has to "have" an identifiable, nameable need if professionals are to provide help. Educational authorities have therefore continued to use categories (or "labels") to classify disabilities in spite of expressing continuing concern about whether the practice hurts students' self-esteem or standing in the eyes of peers (Biklen & Kliewer, 2006). For classroom teachers, the best strategy may be simply to understand how categories of disabilities are defined, while also keeping their limitations in mind and being ready to explain their limitations (tactfully, of course) to parents or others who use the labels inappropriately.
That said, what in fact are the major types of disabilities encountered by teachers? Let us take them one at a time, beginning with the more common ones.
Learning disabilities
A
learning disability
(or
LD
) is a specific impairment of academic learning that interferes with a specific aspect of schoolwork and that reduces a student's academic performance significantly. An LD shows itself as a major discrepancy between a student's ability and some feature of achievement: the student may be delayed in reading, writing, listening, speaking, or doing mathematics, but not in all of these at once. A learning problem is not considered a
learning
disability if it stems from physical, sensory, or motor handicaps, or from generalized intellectual impairment (or mental retardation). It is also not an LD if the learning problem really reflects the challenges of learning English as a second language. Genuine LDs are the learning problems left over after these other possibilities are accounted for or excluded. Typically, a student with an LD has not been helped by teachers' ordinary efforts to assist the student when he or she falls behind academically— though what counts as an "ordinary effort", of course, differs among teachers, schools, and students. Most importantly, though, an LD relates to a fairly specific area of academic learning. A student may be able to read and compute well enough, for example, but not be able to write.
LDs are by far the most common form of special educational need, accounting for half of all students with special needs in the United States and anywhere from 5 to 20 per cent of all students, depending on how the numbers are estimated (United States Department of Education, 2005; Ysseldyke & Bielinski, 2002). Students with LDs are so common, in fact, that most teachers regularly encounter at least one per class in any given school year, regardless of the grade level they teach.
Defining learning disabilities clearly
With so many students defined as having learning disabilities, it is not surprising that the term itself becomes ambiguous in the truest sense of "having many meanings". Specific features of LDs vary considerably. Any of the following students, for example, qualify as having a learning disability, assuming that they have no other disease, condition, or circumstance to account for their behavior:
-
Albert, an eighth-grader, has trouble solving word problems that he reads, but can solve them easily if he hears them orally.
-
Bill, also in eighth grade, has the reverse problem: he can solve word problems only when he can read them, not when he hears them.
-
Carole, a fifth-grader, constantly makes errors when she reads textual material aloud, either leaving out words, adding words, or substituting her own words for the printed text.
-
Emily, in seventh grade, has terrible handwriting; her letters vary in size and wobble all over the page, much like a first- or second-grader.
-
Denny reads very slowly, even though he is in fourth grade. His comprehension suffers as a result, because he sometimes forgets what he read at the beginning of a sentence by the time he reaches the end.
-
Garnet's spelling would have to be called "inventive", even though he has practiced conventionally correct spelling more than other students. Garnet is in sixth grade.
-
Harmin, a ninth-grader has particular trouble decoding individual words and letters if they are unfamiliar; he reads conceal as "concol" and
alternate
as "alfoonite".
-
Irma, a tenth-grader, adds multiple-digit numbers as if they were single-digit numbers stuck together:
42 + 59
equals
911
rather than
101
, though
23 + 54
correctly equals
77
.
With so many expressions of LDs, it is not surprising that educators sometimes disagree about their nature and about the kind of help students need as a consequence. Such controversy may be inevitable because LDs by definition are learning problems with no obvious origin. There is good news, however, from this state of affairs, in that it opens the way to try a variety of solutions for helping students with learning disabilities.
Assisting students with learning disabilities
There are various ways to assist students with learning disabilities, depending not only on the nature of the disability, of course, but also on the concepts or theory of learning guiding you. Take Irma, the girl mentioned above who adds two-digit numbers as if they were one digit numbers. Stated more formally, Irma adds two-digit numbers without carrying digits forward from the ones column to the tens column, or from the tens to the hundreds column. Exhibit 7 shows the effect that her strategy has on one of her homework papers. What is going on here and how could a teacher help Irma?
Exhibit \(\PageIndex{1}\): Irma's math homework about two-digit addition
Directions: Add the following numbers.
|
42
|
23
|
11
|
47
|
97
|
|
+ 59
|
+ 54
|
+ 48
|
+ 23
|
+ 64
|
|
911
|
77
|
59
|
610
|
1511
|
Three out of the six problems are done correctly, even though Irma seems to use an incorrect strategy systematically on all six problems.
Behaviorism: reinforcement for wrong strategies
One possible approach comes from the behaviorist theory discussed in Chapter 2. Irma may persist with the single-digit strategy because it has been reinforced a lot in the past. Maybe she was rewarded so much for adding single-digit numbers {
3+5
,
7+8
etc.) correctly that she generalized this skill to two-digit problems— in fact over generalized it. This explanation is plausible because she would still get many two-digit problems right, as you can confirm by looking at it. In behaviorist terms, her incorrect strategy would still be reinforced, but now only on a "partial schedule of reinforcement". As I pointed out in Chapter 2, partial schedules are especially slow to extinguish, so Irma persists seemingly indefinitely with treating two-digit problems as if they were single-digit problems.
From the point of view of behaviorism, changing Irma's behavior is tricky since the desired behavior (borrowing correctly) rarely happens and therefore cannot be reinforced very often. It might therefore help for the teacher to reward behaviors that compete directly with Irma's inappropriate strategy. The teacher might reduce credit for simply finding the correct answer, for example, and increase credit for a student showing her work— including the work of carrying digits forward correctly. Or the teacher might make a point of discussing Irma's math work with Irma frequently, so as to create more occasions when she can praise Irma for working problems correctly.
Metacognition and responding reflectively
Part of Irma's problem may be that she is thoughtless about doing her math: the minute she sees numbers on a worksheet, she stuffs them into the first arithmetic procedure that comes to mind. Her learning style, that is, seems too impulsive and not reflective enough, as discussed in Chapter 4. Her style also suggests a failure of metacognition (remember that idea from Chapter 2?), which is her self-monitoring of her own thinking and its effectiveness. As a solution, the teacher could encourage Irma to think out loud when she completes two-digit problems— literally get her to "talk her way through" each problem. If participating in these conversations was sometimes impractical, the teacher might also arrange for a skilled classmate to take her place some of the time. Cooperation between Irma and the classmate might help the classmate as well, or even improve overall social relationships in the classroom.
Constructivism, mentoring, and the zone of proximal development
Perhaps Irma has in fact learned how to carry digits forward, but not learned the procedure well enough to use it reliably on her own; so she constantly falls back on the earlier, better-learned strategy of single-digit addition. In that case her problem can be seen in the constructivist terms, like those that I discussed in Chapter 2. In essence, Irma has lacked appropriate mentoring from someone more expert than herself, someone who can create a "zone of proximal development" in which she can display and consolidate her skills more successfully. She still needs mentoring or "assisted coaching" more than independent practice. The teacher can arrange some of this in much the way she encourages to be more reflective, either by working with Irma herself or by arranging for a classmate or even a parent volunteer to do so. In this case, however, whoever serves as mentor should not only listen, but also actively offer Irma help. The help has to be just enough to insure that Irma completes two-digit problems correctly —neither more nor less. Too much help may prevent Irma from taking responsibility for learning the new strategy, but too little may cause her to take the responsibility prematurely.


11.9: Blending Ethos-Pathos-Logos Together
Think about what a speech would look like if a speaker focused solely on one of these three elements.
A speech employing pure ethos, but no pathos or logos, would likely sound charismatic and enjoyable, but upon reconsideration it would appear empty, vapid, and devoid of substance. In the end, audiences would find it hard to make an emotional connection to the topic and likely doubt the information due to the lack of logical arguments made to support the claims.
A speech using pure pathos, but little ethos or logos, would come across as emotionally manipulative.
A speech utilizing pure logos without ethos or pathos would likely sound dry and boring due to the barrage of factual information and lack of emotional or personal connection to relate those claims to the audience.
These points illustrate effectively why Aristotle was so adamant about the need to present balanced arguments. Think of ethos as setting the stage for engaging the audiences to earn their trust and respect while maintaining their attention. When it works in conjunction with ethos, it helps nurture the emotional connection necessary to generate action based on the persuasive messages. Logos effectively balances out pathos by providing substance to those emotional connections and makes good on the promises made by the use of ethos. Using all three elements in careful consideration with one another provides not only a more balanced persuasive message, but a more effective one as well.


10.5: Emotion
Learning Objectives
- Explain the major theories of emotion
- Describe the role that limbic structures play in emotional processing
- Understand the ubiquitous nature of producing and recognizing emotional expression
As we move through our daily lives, we experience a variety of emotions. An emotion is a subjective state of being that we often describe as our feelings. The words emotion and mood are sometimes used interchangeably, but psychologists use these words to refer to two different things. Typically, the word emotion indicates a subjective, affective state that is relatively intense and that occurs in response to something we experience (See figure \(\PageIndex{1}\)). Emotions are often thought to be consciously experienced and intentional. Mood , on the other hand, refers to a prolonged, less intense, affective state that does not occur in response to something we experience. Mood states may not be consciously recognized and do not carry the intentionality that is associated with emotion (Beedie, Terry, Lane, & Devonport, 2011). Here we will focus on emotion, and you will learn more about mood in the chapter that covers psychological disorders.
We can be at the heights of joy or in the depths of despair or. We might feel angry when we are betrayed, fear when we are threatened, and surprised when something unexpected happens. This section will outline some of the most well-known theories explaining our emotional experience and provide insight into the biological bases of emotion. This section closes with a discussion of the ubiquitous nature of facial expressions of emotion and our abilities to recognize those expressions in others.
Theories of Emotion
Our emotional states are combinations of physiological arousal, psychological appraisal, and subjective experiences. Together, these are known as the components of emotion . These appraisals are informed by our experiences, backgrounds, and cultures. Therefore, different people may have different emotional experiences even when faced with similar circumstances. Over time, several different theories of emotion, shown in figure \(\PageIndex{2}\), have been proposed to explain how the various components of emotion interact with one another.
The James-Lange theory of emotion asserts that emotions arise from physiological arousal. Recall what you have learned about the sympathetic nervous system and our fight or flight response when threatened. If you were to encounter some threat in your environment, like a venomous snake in your backyard, your sympathetic nervous system would initiate significant physiological arousal, which would make your heart race and increase your respiration rate. According to the James-Lange theory of emotion, you would only experience a feeling of fear after this physiological arousal had taken place. Furthermore, different arousal patterns would be associated with different feelings.
Other theorists, however, doubted that the physiological arousal that occurs with different types of emotions is distinct enough to result in the wide variety of emotions that we experience. Thus, the Cannon-Bard theory of emotion was developed. According to this view, physiological arousal and emotional experience occur simultaneously, yet independently (Lang, 1994). So, when you see the venomous snake, you feel fear at exactly the same time that your body mounts its fight or flight response. This emotional reaction would be separate and independent of the physiological arousal, even though they co-occur.
The James-Lange and Cannon-Bard theories have each garnered some empirical support in various research paradigms. For instance, Chwalisz, Diener, and Gallagher (1988) conducted a study of the emotional experiences of people who had spinal cord injuries. They reported that individuals who were incapable of receiving autonomic feedback because of their injuries still experienced emotion; however, there was a tendency for people with less awareness of autonomic arousal to experience less intense emotions. More recently, research investigating the facial feedback hypothesis suggested that suppression of facial expression of emotion lowered the intensity of some emotions experienced by participants (Davis, Senghas, & Ochsner, 2009). In both of these examples, neither theory is fully supported because physiological arousal does not seem to be necessary for the emotional experience, but this arousal does appear to be involved in enhancing the intensity of the emotional experience.
The Schachter-Singer two-factor theory of emotion is another variation on theories of emotions that takes into account both physiological arousal and the emotional experience. According to this theory, emotions are composed of two factors: physiological and cognitive. In other words, physiological arousal is interpreted in context to produce the emotional experience. In revisiting our example involving the venomous snake in your backyard, the two-factor theory maintains that the snake elicits sympathetic nervous system activation that is labeled as fear given the context, and our experience is that of fear.
It is important to point out that Schachter and Singer believed that physiological arousal is very similar across the different types of emotions that we experience, and therefore, the cognitive appraisal of the situation is critical to the actual emotion experienced. In fact, it might be possible to misattribute arousal to an emotional experience if the circumstances were right (Schachter & Singer, 1962).
To test their idea, Schachter and Singer performed a clever experiment. Male participants were randomly assigned to one of several groups. Some of the participants received injections of epinephrine that caused bodily changes that mimicked the fight-or-flight response of the sympathetic nervous system; however, only some of these men were told to expect these reactions as side effects of the injection. The other men that received injections of epinephrine were told either that the injection would have no side effects or that it would result in a side effect unrelated to a sympathetic response, such as itching feet or headache. After receiving these injections, participants waited in a room with someone else they thought was another subject in the research project. In reality, the other person was a confederate of the researcher. The confederate engaged in scripted displays of euphoric or angry behavior (Schachter & Singer, 1962).
When those subjects who were told that they should expect to feel symptoms of physiological arousal were asked about any emotional changes that they had experienced related to either euphoria or anger (depending on how their confederate behaved), they reported none. However, the men who weren’t expecting physiological arousal as a function of the injection were more likely to report that they experienced euphoria or anger as a function of their assigned confederate’s behavior. While everyone that received an injection of epinephrine experienced the same physiological arousal, only those who were not expecting the arousal used context to interpret the arousal as a change in emotional state (Schachter & Singer, 1962).
Strong emotional responses are associated with strong physiological arousal. This has led some to suggest that the signs of physiological arousal, which include increased heart rate, respiration rate, and sweating, might serve as a tool to determine whether someone is telling the truth or not. The assumption is that most of us would show signs of physiological arousal if we were being dishonest with someone. A polygraph , or lie detector test, measures the physiological arousal of an individual responding to a series of questions. Someone trained in reading these tests would look for answers to questions that are associated with increased levels of arousal as potential signs that the respondent may have been dishonest on those answers. While polygraphs are still commonly used, their validity and accuracy are highly questionable because there is no evidence that lying is associated with any particular pattern of physiological arousal (Saxe & Ben-Shakhar, 1999).
The relationship between our experiencing of emotions and our cognitive processing of them, and the order in which these occur, remains a topic of research and debate. Lazarus (1991) developed the cognitive-mediational theory that asserts our emotions are determined by our appraisal of the stimulus. This appraisal mediates between the stimulus and the emotional response, and it is immediate and often unconscious. In contrast to the Schachter-Singer model, the appraisal precedes a cognitive label. You will learn more about Lazarus’s appraisal concept when you study stress, health, and lifestyle.
Two other prominent views arise from the work of Robert Zajonc and Joseph LeDoux. Zajonc asserted that some emotions occur separately from or prior to our cognitive interpretation of them, such as feeling fear in response to an unexpected loud sound (Zajonc, 1998). He also believed in what we might casually refer to as a gut feeling—that we can experience an instantaneous and unexplainable like or dislike for someone or something (Zajonc, 1980). LeDoux also views some emotions as requiring no cognition: some emotions completely bypass contextual interpretation. His research into the neuroscience of emotion has demonstrated the amygdala’s primary role in fear (Cunha, Monfils, & LeDoux, 2010; LeDoux 1996, 2002). A fear stimulus is processed by the brain through one of two paths: from the thalamus (where it is perceived) directly to the amygdala or from the thalamus through the cortex and then to the amygdala. The first path is quick, while the second enables more processing about details of the stimulus. In the following section, we will look more closely at the neuroscience of emotional response.
The Biology of Emotions
Earlier, you learned about the limbic system , which is the area of the brain involved in emotion and memory (See figure \(\PageIndex{3}\)). The limbic system includes the hypothalamus, thalamus, amygdala, and the hippocampus. The hypothalamus plays a role in the activation of the sympathetic nervous system that is a part of any given emotional reaction. The thalamus serves as a sensory relay center whose neurons project to both the amygdala and the higher cortical regions for further processing. The amygdala plays a role in processing emotional information and sending that information on to cortical structures (Fossati, 2012).The hippocampus integrates emotional experience with cognition (Femenía, Gómez-Galán, Lindskog, & Magara, 2012).
Amygdala
The amygdala has received a great deal of attention from researchers interested in understanding the biological basis for emotions, especially fear and anxiety (Blackford & Pine, 2012; Goosens & Maren, 2002; Maren, Phan, & Liberzon, 2013). The amygdala is composed of various subnuclei, including the basolateral complex and the central nucleus (See figure \(\PageIndex{4}\)). The basolateral complex has dense connections with a variety of sensory areas of the brain. It is critical for classical conditioning and for attaching emotional value to learning processes and memory. The central nucleus plays a role in attention, and it has connections with the hypothalamus and various brainstem areas to regulate the autonomic nervous and endocrine systems’ activity (Pessoa, 2010).
Animal research has demonstrated that there is increased activation of the amygdala in rat pups that have odor cues paired with electrical shock when their mother is absent. This leads to an aversion to the odor cue that suggests the rats learned to fear the odor cue. Interestingly, when the mother was present, the rats actually showed a preference for the odor cue despite its association with an electrical shock. This preference was associated with no increases in amygdala activation. This suggests a differential effect on the amygdala by the context (the presence or absence of the mother) determined whether the pups learned to fear the odor or to be attracted to it (Moriceau & Sullivan, 2006).
Raineki, Cortés, Belnoue, and Sullivan (2012) demonstrated that, in rats, negative early life experiences could alter the function of the amygdala and result in adolescent patterns of behavior that mimic human mood disorders. In this study, rat pups received either abusive or normal treatment during postnatal days \(8-12\). There were two forms of abusive treatment. The first form of abusive treatment had an insufficient bedding condition.  The second form of abusive treatment had an associative learning task that involved pairing odors and an electrical stimulus in the absence of the mother, as described above. The control group was in a cage with sufficient bedding and was left undisturbed with their mothers during the same time period. The rat pups that experienced abuse were much more likely to exhibit depressive-like symptoms during adolescence when compared to controls. These depressive-like behaviors were associated with increased activation of the amygdala.
Human research also suggests a relationship between the amygdala and psychological disorders of mood or anxiety. Changes in amygdala structure and function have been demonstrated in adolescents who are either at-risk or have been diagnosed with various mood and/or anxiety disorders (Miguel-Hidalgo, 2013; Qin et al., 2013). It has also been suggested that functional differences in the amygdala could serve as a biomarker to differentiate individuals suffering from bipolar disorder from those suffering from major depressive disorder (Fournier, Keener, Almeida, Kronhaus, & Phillips, 2013).
Hippocampus
As mentioned earlier, the hippocampus is also involved in emotional processing. Like the amygdala, research has demonstrated that hippocampal structure and function are linked to a variety of mood and anxiety disorders. Individuals suffering from posttraumatic stress disorder (PTSD) show marked reductions in the volume of several parts of the hippocampus, which may result from decreased levels of neurogenesis and dendritic branching (the generation of new neurons and the generation of new dendrites in existing neurons, respectively) (Wang et al., 2010). While it is impossible to make a causal claim with correlational research like this, studies have demonstrated behavioral improvements and hippocampal volume increases following either pharmacological or cognitive-behavioral therapy in individuals suffering from PTSD (Bremner & Vermetten, 2004; Levy-Gigi, Szabó, Kelemen, & Kéri, 2013).
Facial Expression and Recognition of Emotions
Culture can impact the way in which people display emotion. A cultural display rule is one of a collection of culturally specific standards that govern the types and frequencies of displays of emotions that are acceptable (Malatesta & Haviland, 1982). Therefore, people from varying cultural backgrounds can have very different cultural display rules of emotion. For example, research has shown that individuals from the United States express negative emotions like fear, anger, and disgust both alone and in the presence of others, while Japanese individuals only do so while alone (Matsumoto, 1990). Furthermore, individuals from cultures that tend to emphasize social cohesion are more likely to engage in suppression of emotional reaction so they can evaluate which response is most appropriate in a given context (Matsumoto, Yoo, & Nakagawa, 2008).
Other distinct cultural characteristics might be involved in emotionality. For instance, there may be gender differences involved in emotional processing. While research into gender differences in emotional display is equivocal, there is some evidence that men and women may differ in regulation of emotions (McRae, Ochsner, Mauss, Gabrieli, & Gross, 2008).
Despite different emotional display rules, our ability to recognize and produce facial expressions of emotion appears to be universal. In fact, even congenitally blind individuals produce the same facial expression of emotions, despite their never having the opportunity to observe these facial displays of emotion in other people. This would seem to suggest that the pattern of activity in facial muscles involved in generating emotional expressions is universal, and indeed, this idea was suggested in the late \(19^{th}\) century in Charles Darwin’s book The Expression of Emotions in Man and Animals (1872) . In fact, there is substantial evidence for seven universal emotions that are each associated with distinct facial expressions. These include: happiness, surprise, sadness, fright, disgust, contempt, and anger (See figure below) (Ekman & Keltner, 1997).
Does smiling make you happy? Or does being happy make you smile? The facial feedback hypothesis asserts that facial expressions are capable of influencing our emotions, meaning that smiling can make you feel happier (Buck, 1980; Soussignan, 2001; Strack, Martin, & Stepper, 1988). Recent research explored how Botox, which paralyzes facial muscles and limits facial expression, might affect emotion. Havas, Glenberg, Gutowski, Lucarelli, and Davidson (2010) discovered that depressed individuals reported less depression after paralysis of their frowning muscles with Botox injections.
Of course, emotion is not only displayed through facial expression. We also use the tone of our voices, various behaviors, and body language to communicate information about our emotional states. Body language is the expression of emotion in terms of body position or movement. Research suggests that we are quite sensitive to the emotional information communicated through body language, even if we’re not consciously aware of it (de Gelder, 2006; Tamietto et al., 2009).
CONNECT THE CONCEPTS: Autism Spectrum Disorder and Expression of Emotions
Autism spectrum disorder (ASD) is a set of neurodevelopmental disorders characterized by repetitive behaviors and communication and social problems. Children who have autism spectrum disorders have difficulty recognizing the emotional states of others, and research has shown that this may stem from an inability to distinguish various nonverbal expressions of emotion (i.e., facial expressions) from one another (Hobson, 1986). In addition, there is evidence to suggest that autistic individuals also have difficulty expressing emotion through tone of voice and by producing facial expressions (Macdonald et al., 1989). Difficulties with emotional recognition and expression may contribute to the impaired social interaction and communication that characterize autism; therefore, various therapeutic approaches have been explored to address these difficulties. Various educational curricula, cognitive-behavioral therapies, and pharmacological therapies have shown some promise in helping autistic individuals process emotionally relevant information (Bauminger, 2002; Golan & Baron-Cohen, 2006; Guastella et al., 2010).
Summary
Emotions are subjective experiences that consist of physiological arousal and cognitive appraisal. Various theories have been put forward to explain our emotional experiences. The James-Lange theory asserts that emotions arise as a function of physiological arousal. The Cannon-Bard theory maintains that emotional experience occurs simultaneous to and independent of physiological arousal. The Schachter-Singer two-factor theory suggests that physiological arousal receives cognitive labels as a function of the relevant context and that these two factors together result in an emotional experience.
The limbic system is the brain’s emotional circuit, which includes the amygdala and the hippocampus. Both of these structures are implicated in playing a role in normal emotional processing as well as in psychological mood and anxiety disorders. Increased amygdala activity is associated with learning to fear, and it is seen in individuals who are at risk for or suffering from mood disorders. The volume of the hippocampus has been shown to be reduced in individuals suffering from posttraumatic stress disorder.
The ability to produce and recognize facial expressions of emotions seems to be universal regardless of cultural background. However, there are cultural display rules which influence how often and under what circumstances various emotions can be expressed. Tone of voice and body language also serve as a means by which we communicate information about our emotional states.
Glossary
- basolateral complex
- part of the brain with dense connections with a variety of sensory areas of the brain; it is critical for classical conditioning and attaching emotional value to memory
- body language
- emotional expression through body position or movement
- Cannon-Bard theory of emotion
- physiological arousal and emotional experience occur at the same time
- central nucleus
- part of the brain involved in attention and has connections with the hypothalamus and various brainstem areas to regulate the autonomic nervous and endocrine systems’ activity
- cognitive-mediational theory
- our emotions are determined by our appraisal of the stimulus
- components of emotion
- physiological arousal, psychological appraisal, and subjective experience
- cultural display rule
- one of the culturally specific standards that govern the types and frequencies of emotions that are acceptable
- emotion
- subjective state of being often described as feelings
- facial feedback hypothesis
- facial expressions are capable of influencing our emotions
- James-Lange theory of emotion
- emotions arise from physiological arousal
- polygraph
- lie detector test that measures physiological arousal of individuals as they answer a series of questions
- Schachter-Singer two-factor theory of emotion
- emotions consist of two factors: physiological and cognitive
Contributors and Attributions
-
Rose M. Spielman with many significant contributors. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the creative commons license and may not be reproduced without the prior and express written consent of Rice University. For questions regarding this license, please contact<EMAIL_ADDRESS>.Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>.


3.1: Line styles and types
Standard lines have been developed so that every drawing or sketch conveys the same meaning to everyone. In order to convey that meaning, the lines used in technical drawings have both a definite pattern and a definite thickness. Some lines are complete, and others are broken. Some lines are thick, and others are thin. A visible line, for example, is used to show the edges (or “outline”) of an object and to make it stand out for easy reading.  On the other hand, a center line, which locates the precise center of a hole or shaft, is drawn thin and made with long and short dashes. This makes it easily distinguishable from the visible line.
When you draw, use a fairly sharp pencil of the correct grade and try to maintain an even, consistent pressure to make it easier for you to produce acceptable lines (Figure \(\PageIndex{1}\)). Study the line thicknesses (or “line weights”) shown in Figure \(\PageIndex{2}\) and practice making them.
In computer drafting, the line shape remains the same, but line thickness may not vary as it does in manually created drawings. Some lines, such as center lines, may not cross in the same manner as in a manual drawing. For most computer drafting, line thickness is not important.
| Type | Weight | Line | Description |
|---|---|---|---|
|
Object line
Margin line |
Heavy | Solid line to show visible shapes, edges, and outlines. | |
| Hidden body line | Medium | Broken lines of long and short dashes to show hidden object lines not visible to the eye. | |
| Phantom line | Light | Broken line of short dashes to show alternate positions or movement of a part. | |
| Section line | Light | Unbroken lines arranged in a pattern, usually straight and at a 45° diagonal. | |
| Projection line | Light |
Unbroken lines that extend away from the object or feature for emphasis. |
|
| center line | Light |
Broken lines of long and short dashes to show the center of an object. |
|
|
Extension line/
Dimension line |
Light | Extension lines are small lines that extend outward from an object or feature. Dimension lines span between the extension lines with arrowheads and a given dimension. | |
| Leader line | Light |
Unbroken line usually drawn at an angle, often with a “dogleg” and an arrowhead. A dot is used in place
of an arrowhead where a surface is referenced. Usually accompanied by a label. |
|
|
Cutting plane
line |
Heavy | Broken line of one long and two short dashes to show an imaginary cross-section. The arrowheads show the direction from where the cross-section is viewed. A corresponding image will show the view of A. | |
|
Break lines for
wood and metal |
Heavy | Unbroken freehand or straight zig-zag lines to abbreviate longer spans of wood or metal. | |
|
Break lines for
piping |
Heavy | Curled lines to abbreviate a longer span of pipe. |
Figure \(\PageIndex{2}\): Weight of lines
To properly read and interpret drawings, you must know the meaning of each line and understand how each is used to construct a drawing. The ten most common are often referred to as the “alphabet of lines.” Let’s look at an explanation and example of each type.
Object lines
Object lines (Figure \(\PageIndex{3}\)) are the most common lines used in drawings. These thick, solid lines show the visible edges, corners, and surfaces of a part. Object lines stand out on the drawing and clearly define the outline and features of the object.
Hidden lines
Hidden lines (Figure \(\PageIndex{4}\)) are used to show edges and surfaces that are not visible in a view. These lines are drawn as thin, evenly-spaced dashes. A surface or edge that is shown in one view with an object line will be shown in another view with a hidden line.
Center lines
Center lines (Figure \(\PageIndex{5}\)) are used in drawings for several different applications. The meaning of a center line is normally determined by how it is used. center lines are thin, alternating long and short dashes that are generally used to show hole centers and center positions of rounded features, such as arcs and radii. Arcs are sections of a circle, and radii are rounded corners or edges of a part. center lines can also show the symmetry of an object.
Dimension and extension lines
Dimension and extension lines (Figure \(\PageIndex{6}\)) are thin, solid lines that show the direction, length, and limits of the dimensions of a part. Dimension lines are drawn with an arrowhead at both ends.
Extension lines are drawn close to, but never touching, the edges or surface they limit. They should be perpendicular, or at right angles, to the dimension line. The length of extension lines is generally suited to the number of dimensions they limit.
Leader lines
Leader lines (Figure \(\PageIndex{7}\)) show information such as dimensional notes, material specifications, and process notes. These lines are normally drawn as thin, solid lines with an arrowhead at one end. They are bent or angled at the start but should always end horizontally at the notation. When leader lines reference a surface, a dot is used instead of an arrowhead.
Note that the symbol ø is used to indicate a diameter rather than the abbreviation “DIA.” The number that immediately follows this symbol is the diameter of the hole, followed by the number of holes that must be drilled to that dimension.
Phantom lines
Like center lines, phantom lines (Figure \(\PageIndex{8}\)) are used for several purposes in blueprints. Phantom lines are used to show alternate positions for moving parts and the positions of related or adjacent parts and to eliminate repeated details. Phantom lines are drawn as thin, alternating long dashes separated by two short dashes.
Cutting plane lines
Cutting plane lines (Figure \(\PageIndex{9}\)) show the location and path of imaginary cuts made through parts to show internal details. In most cases, sectional views (or views that show complicated internal details of a part) are indicated by using a cutting plane line. These lines are thick, alternating long lines separated by two short dashes. The arrowheads at each end show the viewing direction of the related sectional view. The two main types of cutting plane lines are straight and offset.
Section lines
Section lines, also known as sectional lining, (Figure \(\PageIndex{10}\)) indicate the surfaces in a sectional view as they would appear if the part were actually cut along the cutting plane line. These are solid lines that are normally drawn at 45-degree angles. Different symbols are used to represent different types of materials.
Break lines
Break lines are drawn to show that a part has been shortened to reduce its size on the drawing. The two variations of break lines common to blueprints are the long break line and the short break line (Figure \(\PageIndex{11}\)). Long break lines are thin solid lines that have zigzags to indicate a break. Short break lines are thick, wavy solid lines that are drawn freehand. When either of these break lines is used to shorten an object, you can assume that the section removed from the part is identical to the portions shown on either side of the break.


7.5: American Trial Courts and the Principle of Orality
principle of orality requires that the trier of fact (generally the jury, but the judge when the defendant waives a jury trial) consider only the evidence that was developed, presented, and received into the record during the trial. As such, jurors should only make their decision based upon the testimony they heard at trial in addition to the documents and physical evidence introduced and admitted by the court. The principle of orality would be violated if, for example, during deliberations, the jury searched the Internet to find information on the defendant or witnesses. Similarly, if the police question the defendant and write a report, the jury cannot consider the contents of the report unless it has been offered in a way that complies with the rules of evidence and the court has received it during the trial. The principle of orality distinguishes the functions of a trial court, developing the evidence, and the function of the appellate courts, reviewing the record for legal error.
adversarial system generally followed by the United States and the inquisitorial system generally followed in most other countries. Frequently in civil law countries (for example, most European nations), the police, prosecutors, or investigating magistrates question witnesses prior to trial and write summaries of their statements called a dossier . In determining guilt, the trier of fact is presented with just the summaries of the witness statements. The trial in civil law countries is less about the presentation of evidence establishing the defendant’s guilt and more about the defendant’s presentation of mitigation evidence which assists the court in giving an appropriate sentence , or sanction.


4.3: The Stages of Policy Development
Dangerous Myths about Juvenile Sex Offenders
Identifying the Problem and Agenda Setting
Formulation and Adoption
Implementation of the Policy
Evaluation
Impact (outcome) evaluations focus on what changes after the introduction of the crime policy. [1] Changes in police patrol practices aimed at reducing the level of residential burglaries in an area are evaluated in terms of subsequent burglaries. The difficulty with impact evaluations is that changes in the crime rate are rarely, if ever, due to a single intervening variable. For example, after the implementation of curfew laws for juvenile offenders, juvenile crime decreased. Can we say that was because of curfew laws? The entire crime rate for America decreased at the same time. Attributing a single outcome based on a solitary intervention is problematic.
Process evaluations consider the implementation of a policy or program and involve determining the procedure used to implement the policy. These are detailed, descriptive accounts of the implementation of the policy including the goals of the program, who is involved, the level of training, the number of clients served, and changes to the program over time. [2] Unfortunately, process evaluations do not address the actual impact policy has on the crime problem, just what was done about a specific issue or who was involved. While this is indeed a limitation, it is essential to know the inner workings of a program or policy if it is to be replicated.
Cost-benefit evaluations , or analysis, seeks to determine if the costs of a policy are justified by the benefits accrued. A ubiquitous example of this would be an evaluation of the popular anti-drug D.A.R.E. program of the 1980s and 1990s. The D.A.R.E. program was a school-based prevention program aimed at preventing drug use among elementary school-aged children. Rigorous evaluations of the program show that it was ineffective and sometimes actually increased drug use in some youth. The cost of this program was roughly $1.3 billion dollars a year (about $173 to $268 per student per year) to implement nationwide (once all related expenses, such as police officer training and services, materials and supplies, school resources, etc., were factored in). [3] Using a cost-benefit analysis, is that a good use of money to support an ineffective program?
- How can this be instituted? Fine the parent? Sentence the parents to jail time? The policy needs to be a concrete solution to a problem. Many states use fines instead of jailing the parents. (Who’s to watch over the children if the parents are locked up?) Fines sound great. This will make sure parents take an active interest in their children because they do not want to have to pay money if their kid gets into trouble.
- Who needs to be involved in lobbying for this law? Legislators? Senators? Local police? Maybe even city officials, local school boards, and religious organizations. So it’s put on the agenda and gets moved onto a ballot for an official vote. The citizens who think their city needs to be tough on crime vote to approve this policy.
- Bam, it’s law. It is implemented and now parents of juveniles delinquents are charged fines. This actually is a law in nearly every state. In the 1990s, Silverton, Oregon, was a model for communities interested in imposing ordinances that hold parents accountable for their children’s behavior. In Silverton, parents can be fined up to $1,000 if their child is found carrying a gun, smoking cigarettes, or using illegal drugs. Parents who agree to attend parenting classes can avoid fines. Within the first two months after the law was passed in early 1995, seven parents were fined and many others registered for parenting classes.
Oregon has ORS 30.761 (2017), which states:
increases recidivism! It’s true! A study of 1,167 youth in Pennsylvania found that the total amount of fines, fees and/or restitution significantly increased the likelihood of recidivism [4] . Justice system–imposed financial penalties increase the likelihood of recidivism in a sample of adolescent offenders [5] In particular, males, non-whites, and youth with prior dispositions and adjudicated with a drug or property offense were at an increased likelihood of recidivism associated with owing fines and fees (Piquero and Jennings, 2016). This is problematic as fees not only increase recidivism but also increase the likelihood of a “revolving door” juvenile justice system for minority youth.
-
Lab, S. (2016).
Crime Prevention: Approaches, Practices, and Evaluations
(9th ed.). New York, NY: Routledge.
↵
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Lab, S. (2016).
Crime Prevention: Approaches, Practices, and Evaluations
(9th ed.). New York, NY: Routledge.
↵
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Shepard, E. (Winter 2001-2002) A new study finds. We wasted billions on D.A.R.E.
Reconsider Quarterly,
↵
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Piquero and Jennings, 2016, Piquero, A. and Jennings, W.G. (2015)
↵
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Youth Violence and Juvenile Justice, 15 (3) p. 325-340).
↵


15.24D: Tularemia
- Recall the mode of transmission and symptoms associated with tularemia
Tularemia (also known as Pahvant Valley plague, rabbit fever, deer fly fever, and Ohara’s fever) is a serious infectious disease caused by the bacterium Francisella tularensis . A Gram-negative, nonmotile coccobacillus, the bacterium has several subspecies with varying degrees of virulence.
The most important of these is F. tularensis tularensis (Type A), which is found in lagomorphs (rabbits and similar animals) in North America, and it is highly virulent in humans and domestic rabbits. F. tularensis palaearctica (Type B) occurs mainly in aquatic rodents (beavers, muskrats) in North America and in hares and small rodents in northern Eurasia. It is less virulent for humans and rabbits.
The primary vectors are ticks and deer flies, but the disease can also be spread through other arthropods. The disease is named after Tulare County, California and most commonly occurs in North America and parts of Europe and Asia. Although outbreaks can occur in the United States, they are rare.
Depending on the site of infection, tularemia has six characteristic clinical symptoms: ulceroglandular, glandular, oropharyngeal, pneumonic, oculoglandular, and typhoidal. The incubation period for tularemia is one to 14 days; most human infections become apparent after three to five days.
In most susceptible mammals, the clinical signs include fever, lethargy, anorexia, signs of septicemia, and possibly, death. Fever is moderate or very high. Tularemia bacilli can be isolated from blood cultures at this stage. The face and eyes redden and become inflamed. Inflammation spreads to the lymph nodes, which enlarge and may suppurate (mimicking bubonic plague), accompanied by a high fever. Death occurs in less than 1% if therapy is initiated promptly.
Francisella tularensis is an intracellular bacterium, meaning it is able to live as a parasite within host cells. It primarily infects macrophages and is able to evade the immune system. The course of disease involves the spread of the organism to multiple organ systems, including the lungs, liver, spleen, and lymphatic system; and differs according to the route of exposure.
Tularemia is primarily treated with streptomycin but can also be treated with gentamicin for ten days and tetracycline-class drugs such as doxycycline for two to three weeks, chloramphenicol, or fluoroquinolones. An attenuated, live vaccine is available, but its use is only for high-risk groups.
Key Points
- Tularemia is an infection common in wild rodents that is passed to humans through contact with infected animal tissues or by ticks, biting flies, and mosquitoes.
- Symptoms vary depending upon the site of infection. The main clinical signs include fever, lethargy, anorexia, muscle pains and signs of septicemia.
- Although tularemia can be life-threatening, most infections can be treated successfully with antibiotics.
Key Terms
- Francisella tularensis : Francisella tularensis is a pathogenic species of gram-negative bacteria and the causative agent of tularemia or rabbit fever.
- macrophage : A white blood cell that phagocytizes necrotic cell debris and foreign material, including viruses, bacteria, and tattoo ink. It presents foreign antigens on MHC II to lymphocytes. Part of the innate immune system.
- tularemia : An infectious disease caused by the bacterium Francisella tularensis.


1.4: Solving Polynomial Equations
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Learning Objectives
By the end of this section, you will be able to:
- Use the Zero Product Property
- Solve quadratic equations by factoring
- Solve equations with polynomial functions
- Solve applications modeled by polynomial equations
Before you get started, take this readiness quiz.
We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.
A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.
POLYNOMIAL EQUATION
A polynomial equation is an equation that contains a polynomial expression.
The degree of the polynomial equation is the degree of the polynomial.
We have already solved polynomial equations of degree one . Polynomial equations of degree one are linear equations are of the form \(ax+b=c\).
We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:
\[x^2+5x+6=0 \qquad 3y^2+4y=10 \qquad 64u^2−81=0 \qquad n(n+1)=42 \nonumber\]
The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get \(n^2+n\).
The general form of a quadratic equation is \(ax^2+bx+c=0\), with \(a\neq 0\). (If \(a=0\), then \(0·x^2=0\) and we are left with no quadratic term.)
QUADRATIC EQUATION
An equation of the form \(ax^2+bx+c=0\) is called a quadratic equation.
\[a,b,\text{ and }c\text{ are real numbers and }a\neq 0\nonumber\]
To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.
Use the Zero Product Property
We will first solve some quadratic equations by using the Zero Product Property . The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.
ZERO PRODUCT PROPERTY
If \(a·b=0\), then either \(a=0\) or \(b=0\) or both.
We will now use the Zero Product Property, to solve a quadratic equation .
Example \(\PageIndex{1}\): How to Solve a Quadratic Equation Using the Zero Product Property
Solve: \((5n−2)(6n−1)=0\).
- Answer
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Example \(\PageIndex{2}\)
Solve: \((3m−2)(2m+1)=0\).
- Answer
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\(m=\frac{2}{3},\space m=−\frac{1}{2}\)
Example \(\PageIndex{3}\)
Solve: \((4p+3)(4p−3)=0\).
- Answer
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\(p=−\frac{3}{4},\space p=\frac{3}{4}\)
USE THE ZERO PRODUCT PROPERTY.
- Set each factor equal to zero.
- Solve the linear equations.
- Check.
Solve Quadratic Equations by Factoring
The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we be sure to start with the quadratic equation in standard form , \(ax^2+bx+c=0\). Then we factor the expression on the left.
Solve: \(2y^2=13y+45\).
- Answer
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Example \(\PageIndex{5}\)
Solve: \(3c^2=10c−8\).
- Answer
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\(c=2,\space c=\frac{4}{3}\)
Example \(\PageIndex{6}\)
Solve: \(2d^2−5d=3\).
- Answer
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\(d=3,\space d=−12\)
SOLVE A QUADRATIC EQUATION BY FACTORING.
- Write the quadratic equation in standard form, \(ax^2+bx+c=0\).
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check. Substitute each solution separately into the original equation.
Before we factor, we must make sure the quadratic equation is in standard form .
Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?
Example \(\PageIndex{7}\)
Solve: \(169q^2=49\).
- Answer
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\(\begin{array} {ll} &169x^2=49 \\ \text{Write the quadratic equation in standard form.} &169x^2−49=0 \\ \text{Factor. It is a difference of squares.} &(13x−7)(13x+7)=0 \\ \text{Use the Zero Product Property to set each factor to }0. & \\ \text{Solve each equation.} &\begin{array} {ll} 13x−7=0 &13x+7=0 \\ 13x=7 &13x=−7 \\ x=\frac{7}{13} &x=−\frac{7}{13} \end{array} \end{array}\)
Check:
We leave the check up to you.
Example \(\PageIndex{8}\)
Solve: \(25p^2=49\).
- Answer
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\(p=\frac{7}{5},p=−\frac{7}{5}\)
Example \(\PageIndex{9}\)
Solve: \(36x^2=121\).
- Answer
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\(x=\frac{11}{6},x=−\frac{11}{6}\)
In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property , one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.
Example \(\PageIndex{10}\)
Solve: \((3x−8)(x−1)=3x\).
- Answer
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\(\begin{array} {ll} &(3x−8)(x−1)=3x \\ \text{Multiply the binomials.} &3x^2−11x+8=3x \\ \text{Write the quadratic equation in standard form.} &3x^2−14x+8=0 \\ \text{Factor the trinomial.} &(3x−2)(x−4)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {ll} 3x−2=0 &x−4=0 \\ 3x=2 &x=4 \\ x=\frac{2}{3} & \end{array} \\ \text{Check your answers.} &\text{The check is left to you.} \end{array}\)
Example \(\PageIndex{11}\)
Solve: \((2m+1)(m+3)=12m\).
- Answer
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\(m=1,\space m=\frac{3}{2}\)
Example \(\PageIndex{12}\)
Solve: \((k+1)(k−1)=8\).
- Answer
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\(k=3,\space k=−3\)
In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.
Example \(\PageIndex{13}\)
Solve: \(3x^2=12x+63\).
- Answer
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\(\begin{array} {ll} &3x^2=12x+63 \\ \text{Write the quadratic equation in standard form.} &3x^2−12x−63=0 \\ \text{Factor the greatest common factor first.} &3(x^2−4x−21)=0 \\ \text{Factor the trinomial.} &3(x−7)(x+3)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {lll} 3\neq 0 &x−7=0 &x+3=0 \\ 3\neq 0 &x=7 &x=−3 \end{array} \\ \text{Check your answers.} &\text{The check is left to you.} \end{array}\)
Example \(\PageIndex{14}\)
Solve: \(18a^2−30=−33a\).
- Answer
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\(a=−\frac{5}{2},a=\frac{2}{3}\)
Example \(\PageIndex{15}\)
Solve: \(123b=−6−60b^2\)
- Answer
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\(b=−2,\space b=−\frac{1}{20}\)
The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.
Example \(\PageIndex{16}\)
Solve: \(9m^3+100m=60m^2\)
- Answer
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\(\begin{array} {ll} & 9m^3+100m=60m^2 \\ \text{Bring all the terms to one side so that the other side is zero.} &9m^3−60m^2+100m=0 \\ \text{Factor the greatest common factor first.} &m(9m^2−60m+100)=0 \\ \text{Factor the trinomial.} &m(3m−10)^2=0 \end{array}\\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} &\begin{array} {lll} m=0 &3m−10=0 &{}\\ m=0 &m=\frac{10}{3} & {} \end{array}\\ \text{Check your answers.} &\text{The check is left to you.} \end{array}\)
Example \(\PageIndex{17}\)
Solve: \(8x^3=24x^2−18x\).
- Answer
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\(x=0,\space x=\frac{3}{2}\)
Example \(\PageIndex{18}\)
Solve: \(16y^2=32y^3+2y\).
- Answer
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\(y=0,\space y=14\)
Solve Equations with Polynomial Functions
As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.
Example \(\PageIndex{19}\)
For the function \(f(x)=x^2+2x−2\),
ⓐ find \(x\) when \(f(x)=6\)
ⓑ find two points that lie on the graph of the function.
- Answer
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ⓐ
\(\begin{array} {ll} &f(x)=x^2+2x−2 \\ \text{Substitute }6\text{ for }f(x). &6=x^2+2x−2 \\ \text{Put the quadratic in standard form.} &x^2+2x−8=0 \\ \text{Factor the trinomial.} &(x+4)(x−2)=0 \\ \begin{array} {l} \text{Use the zero product property.} \\ \text{Solve.} \end{array} &\begin{array} {lll} x+4=0 &\text{or} &x−2=0 \\ x=−4 &\text{or} &x=2 \end{array} \\ \text{Check:} & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ \begin{array} {lll} \quad &\hspace{3mm} f(x)=x^2+2x−2 &f(x)=x^2+2x−2 \\ \quad &f(−4)=(−4)^2+2(−4)−2 &f(2)=2^2+2·2−2 \\ \quad &f(−4)=16−8−2 &f(2)=4+4−2 \\ \quad &f(−4)=6\checkmark &f(2)=6\checkmark \end{array} & \end{array} \)
ⓑ Since \(f(−4)=6\) and \(f(2)=6\), the points \((−4,6)\) and \((2,6)\) lie on the graph of the function.
Example \(\PageIndex{20}\)
For the function \(f(x)=x^2−2x−8\),
ⓐ find \(x\) when \(f(x)=7\)
ⓑ Find two points that lie on the graph of the function.
- Answer
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ⓐ \(x=−3\) or \(x=5\)
ⓑ \((−3,7)\space (5,7)\)
Example \(\PageIndex{21}\)
For the function \(f(x)=x^2−8x+3\),
ⓐ find \(x\) when \(f(x)=−4\)
ⓑ Find two points that lie on the graph of the function.
- Answer
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ⓐ \(x=1\) or \(x=7\)
ⓑ \((1,−4)\space (7,−4)\)
The Zero Product Property also helps us determine where the function is zero. A value of \(x\) where the function is \(0\), is called a zero of the function .
ZERO OF A FUNCTION
For any function \(f\), if \(f(x)=0\), then \(x\) is a zero of the function .
When \(f(x)=0\), the point \((x,0)\) is a point on the graph. This point is an \(x\)- intercept of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later.
Example \(\PageIndex{22}\)
For the function \(f(x)=3x^2+10x−8\), find
ⓐ the zeros of the function,
ⓑ any \(x\)-intercepts of the graph of the function
ⓒ any \(y\)-intercepts of the graph of the function
- Answer
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ⓐ To find the zeros of the function, we need to find when the function value is 0.
\(\begin{array} {ll} &f(x)=3x^2+10x−8 \\ \text{Substitute }0\text{ for}f(x). &0=3x^2+10x−8 \\ \text{Factor the trinomial.} &(x+4)(3x−2)=0 \\ \begin{array} {l} \text{Use the zero product property.} \\ \text{Solve.} \end{array} &\begin{array} {lll} x+4=0 &\text{or} &3x−2=0 \\ x=−4 &\text{or} &x=\frac{2}{3} \end{array} \end{array}\)
ⓑ An \(x\)-intercept occurs when \(y=0\). Since \(f(−4)=0\) and \(f(\frac{2}{3})=0\), the points \((−4,0)\) and \((\frac{2}{3},0)\) lie on the graph. These points are \(x\)-intercepts of the function.
ⓒ A \(y\)-intercept occurs when \(x=0\). To find the \(y\)-intercepts we need to find \(f(0)\).
\(\begin{array} {ll} &f(x)=3x^2+10x−8 \\ \text{Find }f(0)\text{ by substituting }0\text{ for }x. &f(0)=3·0^2+10·0−8 \\ \text{Simplify.} &f(0)=−8 \end{array} \)
Since \(f(0)=−8\), the point \((0,−8)\) lies on the graph. This point is the \(y\)-intercept of the function.
Example \(\PageIndex{23}\)
For the function \(f(x)=2x^2−7x+5\), find
ⓐ the zeros of the function
ⓑ any \(x\)-intercepts of the graph of the function
ⓒ any \(y\)-intercepts of the graph of the function.
- Answer
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ⓐ \(x=1\) or \(x=\frac{5}{2}\)
ⓑ \((1,0),\space (\frac{5}{2},0)\) ⓒ \((0,5)\)
Example \(\PageIndex{24}\)
For the function \(f(x)=6x^2+13x−15\), find
ⓐ the zeros of the function
ⓑ any \(x\)-intercepts of the graph of the function
ⓒ any \(y\)-intercepts of the graph of the function.
- Answer
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ⓐ \(x=−3\) or \(x=\frac{5}{6}\)
ⓑ \((−3,0),\space (\frac{5}{6},0)\) ⓒ \((0,−15)\)
Solve Applications Modeled by Polynomial Equations
The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.
USE A PROBLEM SOLVING STRATEGY TO SOLVE WORD PROBLEMS.
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
- Solve the equation using appropriate algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
We will start with a number problem to get practice translating words into a polynomial equation.
Example \(\PageIndex{25}\)
The product of two consecutive odd integers is 323. Find the integers.
- Answer
-
\(\begin{array} {ll} \textbf{Step 1. Read }\text{the problem.} & \\ \textbf{Step 2. Identify }\text{what we are looking for.} &\text{We are looking for two consecutive integers.} \\ \textbf{Step 3. Name}\text{ what we are looking for.} &\text{Let } n=\text{ the first integer.} \\ &n+2= \text{ next consecutive odd integer} \\ \begin{array} {l} \textbf{Step 4. Translate }\text{into an equation. Restate the}\hspace{20mm} \\ \text{problem in a sentence.} \end{array} &\begin{array} {l} \text{The product of the two consecutive odd} \\ \text{integers is }323. \end{array} \\ &\quad n(n+2)=323 \\ \textbf{Step 5. Solve }\text{the equation.} n^2+2n=323 \\ \text{Bring all the terms to one side.} &n^2+2n−323=0 \\ \text{Factor the trinomial.} &(n−17)(n+19)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve the equations.} \end{array} &\begin{array} {ll} n−17=0 \hspace{10mm}&n+19=0 \\ n=17 &n=−19 \end{array} \end{array} \)
There are two values for \(n\) that are solutions to this problem. So there are two sets of consecutive odd integers that will work.\(\begin{array} {ll} \text{If the first integer is } n=17 \hspace{60mm} &\text{If the first integer is } n=-19 \\ \text{then the next odd integer is} &\text{then the next odd integer is} \\ \hspace{53mm} n+2 &\hspace{53mm} n+2 \\ \hspace{51mm} 17+2 &\hspace{51mm} -19+2 \\ \hspace{55mm} 19 &\hspace{55mm} -17 \\ \hspace{51mm} 17,19 &\hspace{51mm} -17,-19 \\ \textbf{Step 6. Check }\text{the answer.} & \\ \text{The results are consecutive odd integers} & \\ \begin{array} {ll} 17,\space 19\text{ and }−19,\space −17. & \\ 17·19=323\checkmark &−19(−17)=323\checkmark \end{array} & \\ \text{Both pairs of consecutive integers are solutions.} & \\ \textbf{Step 7. Answer }\text{the question} &\text{The consecutive integers are }17, 19\text{ and }−19,−17. \end{array} \)
Example \(\PageIndex{26}\)
The product of two consecutive odd integers is 255. Find the integers.
- Answer
-
\(−15,−17\) and \(15, 17\)
Example \(\PageIndex{27}\)
The product of two consecutive odd integers is 483 Find the integers.
- Answer
-
\(−23,−21\) and \(21, 23\)
Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.
In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.
Example \(\PageIndex{28}\)
A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.
- Answer
-
Step 1. Read the problem. In problems involving
geometric figures, a sketch can help you visualize
the situation.Step 2. Identify what you are looking for. We are looking for the length and width. Step 3. Name what you are looking for. Let \(w=\text{ the width of the bedroom}\). The length is four feet more than the width. \(w+4=\text{ the length of the garden}\) Step 4. Translate into an equation. Restate the important information in a sentence. The area of the bedroom is 117 square feet. Use the formula for the area of a rectangle. \(A=l·w\) Substitute in the variables. \(117=(w+4)w\) Step 5. Solve the equation Distribute first. \(117=w^2+4w\) Get zero on one side. \(117=w^2+4w\) Factor the trinomial. \(0=w^2+4w−117\) Use the Zero Product Property. \(0=(w^2+13)(w−9)\) Solve each equation. \(0=w+13\quad 0=w−9\) Since \(w\) is the width of the bedroom, it does not
make sense for it to be negative. We eliminate that value for \(w\).\(\cancel{w=−13}\) \(\quad w=9\) \(w=9\) The width is 9 feet. Find the value of the length. \(w+4\)
\(9+4\)
13 The length is 13 feet.Step 6. Check the answer.
Does the answer make sense?
Yes, this makes sense.Step 7. Answer the question. The width of the bedroom is 9 feet and
the length is 13 feet.
Example \(\PageIndex{29}\)
A rectangular sign has area 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.
- Answer
-
The width is 5 feet and length is 6 feet.
Example \(\PageIndex{30}\)
A rectangular patio has area 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.
- Answer
-
The length of the patio is 12 feet and the width 15 feet.
In the next example, we will use the Pythagorean Theorem \((a^2+b^2=c^2)\). This formula gives the relation between the legs and the hypotenuse of a right triangle.
We will use this formula to in the next example.
Example \(\PageIndex{31}\)
A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail.
- Answer
-
Step 1. Read the problem Step 2. Identify what you are looking for. We are looking for the lengths of the
sides of the sail.Step 3. Name what you are looking for.
One side is 7 less than the other.Let \(x=\text{ length of a side of the sail}\).
\(x−7=\text{ length of other side}\)Step 4. Translate into an equation. Since this is a
right triangle we can use the Pythagorean Theorem.\(a^2+b^2=c^2\) Substitute in the variables. \(x^2+(x−7)^2=17^2\) Step 5. Solve the equation
Simplify.\(x^2+x^2−14x+49=289\) \(2x^2−14x+49=289\) It is a quadratic equation, so get zero on one side. \(2x^2−14x−240=0\) Factor the greatest common factor. \(2(x^2−7x−120)=0\) Factor the trinomial. \(2(x−15)(x+8)=0\) Use the Zero Product Property. \(2\neq 0\quad x−15=0\quad x+8=0\) Solve. \(2\neq 0\quad x=15\quad x=−8\) Since \(x\) is a side of the triangle, \(x=−8\) does not
make sense.\(2\neq 0\quad x=15\quad \cancel{x=−8}\) Find the length of the other side. If the length of one side is
then the length of the other side is
8 is the length of the other side.Step 6. Check the answer in the problem
Do these numbers make sense?
Step 7. Answer the question The sides of the sail are 8, 15 and 17 feet.
Example \(\PageIndex{32}\)
Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.
- Answer
-
5 feet and 12 feet
Example \(\PageIndex{33}\)
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.
- Answer
-
24 feet and 25 feet
The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.
Example \(\PageIndex{34}\)
Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function \(h(t)=−16t^2+64t+80\) models the height, \(h\), of the ball above the ground as a function of time, \(t\). Find:
ⓐ the zeros of this function which tell us when the ball hits the ground
ⓑ when the ball will be 80 feet above the ground
ⓒ the height of the ball at \(t=2\) seconds.
- Answer
-
ⓐ The zeros of this function are found by solving \(h(t)=0\). This will tell us when the ball will hit the ground.
\(\begin{array} {ll} &h(t)=0 \\ \text{Substitute in the polynomial for }h(t). &−16t^2+64t+80=0 \\ \text{Factor the GCF, }−16. &−16(t^2−4t−5)=0 \\ \text{Factor the trinomial.} &−16(t−5)(t+1)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve.} \end{array} &\begin{array} {ll} t−5=0 &t+1=0 \\ t=5 &t=−1 \end{array} \end{array} \)The result \(t=5\) tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result \(t=−1\) is discarded.
ⓑ The ball will be 80 feet above the ground when \(h(t)=80\).
\(\begin{array} {ll} &h(t)=80 \\ \text{Substitute in the polynomial for }h(t). &−16t^2+64t+80=80 \\ \text{Subtract 80 from both sides.} &−16t^2+64t=0 \\ \text{Factor the GCF, }−16t. &−16t(t−4)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve.}\end{array} &\begin{array} {ll} −16t=0 &t−4=0 \\ t=0 &t=4 \end{array} \\ &\text{The ball will be at 80 feet the moment Dennis} \\ &\text{tosses the ball and then 4 seconds later, when} \\ &\text{the ball is falling.} \end{array} \)ⓒ To find the height ball at \(t=2\) seconds we find \(h(2)\).
\(\begin{array} {ll} &h(t)=−16t^2+64t+80 \\ \text{To find }h(2)\text{ substitute }2\text{ for }t. &h(2)=−16(2)^2+64·2+80 \\ \text{Simplify.} &h(2)=144 \\ &\text{After 2 seconds, the ball will be at 144 feet.} \end{array}\)
Example \(\PageIndex{35}\)
Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function \(h(t)=−16t^2+48t+160\) models the height, \(h\), of the rock above the ocean as a function of time, \(t\). Find:
ⓐ the zeros of this function which tell us when the rock will hit the ocean
ⓑ when the rock will be 160 feet above the ocean.
ⓒ the height of the rock at \(t=1.5\) seconds.
- Answer
-
ⓐ 5 ⓑ 0;3 ⓒ 196
Example \(\PageIndex{36}\)
Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function \(h(t)=−16t^2+32t+128\) models the height, \(h\), of the penny above the ocean as a function of time, \(t\). Find:
ⓐ the zeros of this function which is when the penny will hit the ocean
ⓑ when the penny will be 128 feet above the ocean.
ⓒ the height the penny will be at \(t=1\) seconds which is when the penny will be at its highest point.
- Answer
-
ⓐ 4 ⓑ 0;2 ⓒ 144
Access this online resource for additional instruction and practice with quadratic equations.
Key Concepts
- Polynomial Equation: A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.
-
Quadratic Equation:
An equation of the form \(ax^2+bx+c=0\) is called a quadratic equation.
\[a,b,c\text{ are real numbers and } a\neq 0\nonumber\]
- Zero Product Property: If \(a·b=0\), then either \(a=0\) or \(b=0\) or both.
-
How to use the Zero Product Property
- Set each factor equal to zero.
- Solve the linear equations.
- Check.
-
How to solve a quadratic equation by factoring.
- Write the quadratic equation in standard form, \(ax^2+bx+c=0\).
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check. Substitute each solution separately into the original equation.
- Zero of a Function: For any function \(f\), if \(f(x)=0\), then \(x\) is a zero of the function.
-
How to use a problem solving strategy to solve word problems.
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
- Solve the equation using appropriate algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.


3.3: Linear Congruences
Because congruences are analogous to equations, it is natural to ask about solutions of linear equations. In this section, we will be discussing linear congruences of one variable and their solutions. We start by defining linear congruences.
A congruence of the form \(ax\equiv b(mod\ m)\) where \(x\) is an unknown integer is called a linear congruence in one variable.
It is important to know that if \(x_0\) is a solution for a linear congruence, then all integers \(x_i\) such that \(x_i\equiv x_0 (mod \ m)\) are solutions of the linear congruence. Notice also that \(ax\equiv b (mod\ m)\) is equivalent to a linear Diophantine equation i.e. there exists \(y\) such that \(ax-my=b\). We now prove theorems about the solutions of linear congruences.
Let \(a,b\) and \(m\) be integers such that \(m>0\) and let \(c=(a,m)\). If \(c\) does not divide \(b\), then the congruence \(ax\equiv b(mod \ m)\) has no solutions. If \(c\mid b\), then \[ax\equiv b(mod \ m)\] has exactly \(c\) incongruent solutions modulo \(m\).
As we mentioned earlier, \(ax\equiv b(mod \ m)\) is equivalent to \(ax-my=b\). By Theorem 19 on Diophantine equations, we know that if \(c\) does not divide \(b\), then the equation, \(ax-my=b\) has no solutions. Notice also that if \(c\mid b\), then there are infinitely many solutions whose variable \(x\) is given by \[x=x_0+(m/c)t\] Thus the above values of \(x\) are solutions of the congruence \(ax\equiv b(mod \ m)\). Now we have to determine the number of incongruent solutions that we have. Suppose that two solutions are congruent, i.e. \[x_0+(m/c)t_1\equiv x_0+(m/c)t_2(mod \ m).\] Thus we get \[(m/c)t_1\equiv (m/c)t_2(mod \ m).\] Now notice that \((m,m/c)=m/c\) and thus \[t_1\equiv t_2(mod \ c).\] Thus we get a set of incongruent solutions given by \(x=x_0+(m/c)t\), where \(t\) is taken modulo \(c\).
Notice that if \(c=(a,m)=1\), then there is a unique solution modulo m for the equation \(ax\equiv b(mod \ m)\).
Let us find all the solutions of the congruence \(3x\equiv 12 (mod \ 6)\). Notice that \((3,6)=3\) and \(3\mid 12\). Thus there are three incongruent solutions modulo \(6\). We use the Euclidean algorithm to find the solution of the equation \(3x-6y=12\) as described in chapter 2. As a result, we get \(x_0=6\). Thus the three incongruent solutions are given by \(x_1=6(mod \ 6)\), \(x_1=6+2=2(mod \ 6)\) and \(x_2=6+4=4(mod \ 6)\).
As we mentioned earlier in Remark 2, the congruence \(ax\equiv b(mod \ m)\) has a unique solution if \((a,m)=1\). This will allow us to talk about modular inverses.
A solution for the congruence \(ax\equiv 1 (mod\ m)\) for \((a,m)=1\) is called the modular inverse of \(a\) modulo m. We denote such a solution by \(\bar{a}\).
The modular inverse of 7 modulo 48 is 7. Notice that a solution for \(7x\equiv 1(mod \ 48)\) is \(x\equiv 7 (mod \ 48)\).
Exercises
- Find all solutions of \(3x\equiv 6(mod \ 9)\) .
- Find all solutions of \(3x\equiv 2(mod \ 7)\) .
- Find an inverse modulo 13 of 2 and of 11.
- Show that if \(\bar{a}\) is the inverse of \(a\) modulo \(m\) and \(\bar{b}\) is the inverse of \(b\) modulo \(m\), then \(\bar{a}\bar{b}\) is the inverse of \(ab\) modulo \(m\).
Contributors and Attributions
-
Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution ( CC BY ) license.


7.3: Linear Equations
The linear first-order differential equation (linear in \(y\) and its derivative) can be written in the form
\[\frac{d y}{d x}+p(x) y=g(x) \nonumber \]
with the initial condition \(y\left(x_{0}\right)=y_{0}\) . Linear first-order equations can be integrated using an integrating factor \(\mu(x)\) . We multiply Equation \ref{7.8} by \(\mu(x)\) ,
\[\mu(x)\left[\frac{d y}{d x}+p(x) y\right]=\mu(x) g(x) \nonumber \]
and try to determine \(\mu(x)\) so that
\[\mu(x)\left[\frac{d y}{d x}+p(x) y\right]=\frac{d}{d x}[\mu(x) y] \nonumber \]
Equation Equation \ref{7.9} then becomes
\[\frac{d}{d x}[\mu(x) y]=\mu(x) g(x) \nonumber \]
Equation Equation \ref{7.11} is easily integrated using \(\mu\left(x_{0}\right)=\mu_{0}\) and \(y\left(x_{0}\right)=y_{0}\) :
\[\mu(x) y-\mu_{0} y_{0}=\int_{x_{0}}^{x} \mu(x) g(x) d x \nonumber \]
\[y=\frac{1}{\mu(x)}\left(\mu_{0} y_{0}+\int_{x_{0}}^{x} \mu(x) g(x) d x\right) . \nonumber \]
It remains to determine \(\mu(x)\) from Equation \ref{7.10}. Differentiating and expanding Equation \ref{7.10} yields
\[\mu \frac{d y}{d x}+p \mu y=\frac{d \mu}{d x} y+\mu \frac{d y}{d x} ; \nonumber \]
and upon simplifying,
\[\frac{d \mu}{d x}=p \mu . \nonumber \]
Equation Equation \ref{7.13} is separable and can be integrated:
\[\begin{gathered} \int_{\mu_{0}}^{\mu} \frac{d \mu}{\mu}=\int_{x_{0}}^{x} p(x) d x, \\ \ln \frac{\mu}{\mu_{0}}=\int_{x_{0}}^{x} p(x) d x \\ \mu(x)=\mu_{0} \exp \left(\int_{x_{0}}^{x} p(x) d x\right) . \end{gathered} \nonumber \]
Notice that since \(\mu_{0}\) cancels out of \(Equation \ref{7.12}\) , it is customary to assign \(\mu_{0}=1\) . The solution to Equation \ref{7.8} satisfying the initial condition \(y\left(x_{0}\right)=y_{0}\) is then commonly written as
\[y=\frac{1}{\mu(x)}\left(y_{0}+\int_{x_{0}}^{x} \mu(x) g(x) d x\right) \nonumber \]
with
\[\mu(x)=\exp \left(\int_{x_{0}}^{x} p(x) d x\right) \nonumber \]
the integrating factor. This important result finds frequent use in applied mathematics.
Example: Solve \(\frac{d y}{d x}+2 y=e^{-x}\) , with \(y(0)=3 / 4\) .
Note that this equation is not separable. With \(p(x)=2\) and \(g(x)=e^{-x}\) , we have
\[\begin{aligned} \mu(x) &=\exp \left(\int_{0}^{x} 2 d x\right) \\ &=e^{2 x} \end{aligned} \nonumber \]
and
\[\begin{aligned} y &=e^{-2 x}\left(\frac{3}{4}+\int_{0}^{x} e^{2 x} e^{-x} d x\right) \\ &=e^{-2 x}\left(\frac{3}{4}+\int_{0}^{x} e^{x} d x\right) \\ &=e^{-2 x}\left(\frac{3}{4}+\left(e^{x}-1\right)\right) \\ &=e^{-2 x}\left(e^{x}-\frac{1}{4}\right) \\ &=e^{-x}\left(1-\frac{1}{4} e^{-x}\right) \end{aligned} \nonumber \]
Example: Solve \(\frac{d y}{d x}-2 x y=x\) , with \(y(0)=0\) .
This equation is separable, and we solve it in two ways. First, using an integrating factor with \(p(x)=-2 x\) and \(g(x)=x\) :
\[\begin{aligned} \mu(x) &=\exp \left(-2 \int_{0}^{x} x d x\right) \\ &=e^{-x^{2}} \end{aligned} \nonumber \]
and
\[y=e^{x^{2}} \int_{0}^{x} x e^{-x^{2}} d x \nonumber \]
The integral can be done by substitution with \(u=x^{2}, d u=2 x d x\) :
\[\begin{aligned} \int_{0}^{x} x e^{-x^{2}} d x &=\frac{1}{2} \int_{0}^{x^{2}} e^{-u} d u \\ &\left.=-\frac{1}{2} e^{-u}\right]_{0}^{x^{2}} \\ &=\frac{1}{2}\left(1-e^{-x^{2}}\right) \end{aligned} \nonumber \]
Therefore,
\[\begin{aligned} y &=\frac{1}{2} e^{x^{2}}\left(1-e^{-x^{2}}\right) \\ &=\frac{1}{2}\left(e^{x^{2}}-1\right) \end{aligned} \nonumber \]
Second, we integrate by separating variables:
\[\begin{gathered} \frac{d y}{d x}-2 x y=x, \\ \frac{d y}{d x}=x(1+2 y), \\ \int_{0}^{y} \frac{d y}{1+2 y}=\int_{0}^{x} x d x \\ \frac{1}{2} \ln (1+2 y)=\frac{1}{2} x^{2}, \\ 1+2 y=e^{x^{2}} \\ y=\frac{1}{2}\left(e^{x^{2}}-1\right) \end{gathered} \nonumber \]
The results from the two different solution methods are the same, and the choice of method is a personal preference.