[Update] Absolute Value Equations | absolute value – Pickpeup

absolute value: คุณกำลังดูกระทู้

 

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Section 2-14 : Absolute Value Equations

In the final two sections of this chapter we want to discuss solving equations and inequalities that contain absolute values. We will look at equations with absolute value in them in this section and we’ll look at inequalities in the next section.

Before solving however, we should first have a brief discussion of just what absolute value is. The notation for the absolute value of \(p\) is \(\left| p \right|\). Note as well that the absolute value bars are NOT parentheses and, in many cases, don’t behave as parentheses so be careful with them.

There are two ways to define absolute value. There is a geometric definition and a mathematical definition. We will look at both.

Geometric Definition

In this definition we are going to think of \(\left| p \right|\) as the distance of \(p\) from the origin on a number line. Also, we will always use a positive value for distance. Consider the following number line.

From this we can get the following values of absolute value.

\[\left| 2 \right| = 2\hspace{0.25in}\left| { – 3} \right| = 3\hspace{0.25in}\left| {\frac{9}{2}} \right| = \frac{9}{2}\]

\[\left| 2 \right| = 2\hspace{0.25in}\left| { – 3} \right| = 3\hspace{0.25in}\left| {\frac{9}{2}} \right| = \frac{9}{2}\]

All that we need to do is identify the point on the number line and determine its distance from the origin. Note as well that we also have \(\left| 0 \right| = 0\).

Mathematical Definition

We can also give a strict mathematical/formula definition for absolute value. It is,

\[\left| p \right| = \left\{ {\begin{array}{*{20}{l}}p&{{\mbox{if }}p \ge 0}\\{ – p}&{{\mbox{if }}p < 0}\end{array}} \right.\]

\[\left| p \right| = \left\{ {\begin{array}{*{20}{l}}p&{{\mbox{if }}p \ge 0}\\{ – p}&{{\mbox{if }}p < 0}\end{array}} \right.\]

This tells us to look at the sign of \(p\) and if it’s positive we just drop the absolute value bar. If \(p\) is negative we drop the absolute value bars and then put in a negative in front of it.

So, let’s see a couple of quick examples.

\[\begin{align*}\left| 4 \right| & = 4\hspace{0.25in}{\mbox{because }}4 \ge 0\\ \left| { – 8} \right| & = – \left( { – 8} \right) = 8\hspace{0.25in}{\mbox{because }} – 8 < 0\\ \left| 0 \right| & = 0\hspace{0.25in}{\mbox{because }}0 \ge 0\end{align*}\]

\[\begin{align*}\left| 4 \right| & = 4\hspace{0.25in}{\mbox{because }}4 \ge 0\\ \left| { – 8} \right| & = – \left( { – 8} \right) = 8\hspace{0.25in}{\mbox{because }} – 8 < 0\\ \left| 0 \right| & = 0\hspace{0.25in}{\mbox{because }}0 \ge 0\end{align*}\]

Note that these give exactly the same value as if we’d used the geometric interpretation.

One way to think of absolute value is that it takes a number and makes it positive. In fact, we can guarantee that,

\[\left| p \right| \ge 0\]

\[\left| p \right| \ge 0\]

regardless of the value of \(p\).

We do need to be careful however to not misuse either of these definitions. For example, we can’t use the definition on

\[\left| { – x} \right|\]

\[\left| { – x} \right|\]

because we don’t know the value of \(x\).

Also, don’t make the mistake of assuming that absolute value just makes all minus signs into plus signs. In other words, don’t make the following mistake,

\[\left| {4x – 3} \right| \ne 4x + 3\]

\[\left| {4x – 3} \right| \ne 4x + 3\]

This just isn’t true! If you aren’t sure that you believe that then plug in a number for \(x\). For example, if \(x = – 1\) we would get,

\[7 = \left| { – 7} \right| = \left| {4\left( { – 1} \right) – 3} \right| \ne 4\left( { – 1} \right) + 3 = – 1\]

\[7 = \left| { – 7} \right| = \left| {4\left( { – 1} \right) – 3} \right| \ne 4\left( { – 1} \right) + 3 = – 1\]

There are a couple of problems with this. First, the numbers are clearly not the same and so that’s all we really need to prove that the two expressions aren’t the same. There is also the fact however that the right number is negative and we will never get a negative value out of an absolute value! That also will guarantee that these two expressions aren’t the same.

The definitions above are easy to apply if all we’ve got are numbers inside the absolute value bars. However, once we put variables inside them we’ve got to start being very careful.

It’s now time to start thinking about how to solve equations that contain absolute values. Let’s start off fairly simple and look at the following equation.

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\[\left| p \right| = 4\]

\[\left| p \right| = 4\]

Now, if we think of this from a geometric point of view this means that whatever \(p\) is it must have a distance of 4 from the origin. Well there are only two numbers that have a distance of 4 from the origin, namely 4 and -4. So, there are two solutions to this equation,

\[p = – 4\hspace{0.25in}{\mbox{or}}\hspace{0.25in}p = 4\]

\[p = – 4\hspace{0.25in}{\mbox{or}}\hspace{0.25in}p = 4\]

Now, if you think about it we can do this for any positive number, not just 4. So, this leads to the following general formula for equations involving absolute value.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{If }}\left| p \right| = b,\,\,\,b > 0\,\,\,\,\,\,\,\,\,\,\,{\mbox{then }}p = – b\,\,\,{\mbox{or }}p = b}}\]

Notice that this does require the \(b\) be a positive number. We will deal with what happens if \(b\) is zero or negative in a bit.

Let’s take a look at some examples.

Example 1

Solve each of the following.

  1. \(\left| {2x – 5} \right| = 9\)
  2. \(\left| {1 – 3t} \right| = 20\)
  3. \(\left| {5y – 8} \right| = 1\)

 

Now, remember that absolute value does not just make all minus signs into plus signs! To solve these, we’ve got to use the formula above since in all cases the number on the right side of the equal sign is positive.

a

\(\left| {2x – 5} \right| = 9\)

There really isn’t much to do here other than using the formula from above as noted above. All we need to note is that in the formula above \(p\) represents whatever is on the inside of the absolute value bars and so in this case we have,

\[2x – 5 = – 9\hspace{0.25in}{\mbox{or}}\hspace{0.25in}2x – 5 = 9\]

\[2x – 5 = – 9\hspace{0.25in}{\mbox{or}}\hspace{0.25in}2x – 5 = 9\]

At this point we’ve got two linear equations that are easy to solve.

\[\begin{align*}2x & = – 4\hspace{0.25in}{\mbox{or}}\hspace{0.25in}2x = 14\\ x & = – 2\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\,\,\,x = 7\end{align*}\]

\[\begin{align*}2x & = – 4\hspace{0.25in}{\mbox{or}}\hspace{0.25in}2x = 14\\ x & = – 2\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\,\,\,x = 7\end{align*}\]

So, we’ve got two solutions to the equation \(x = – 2\) and \(x = 7\).

b

\(\left| {1 – 3t} \right| = 20\)

This one is pretty much the same as the previous part so we won’t put as much detail into this one.

\[\begin{align*}1 – 3t & = – 20\hspace{0.25in}{\mbox{or}}\hspace{0.25in}1 – 3t = 20\\ – 3t & = – 21\hspace{0.25in}{\mbox{or}}\hspace{0.25in} – 3t = 19\\ t & = 7\hspace{0.25in}{\mbox{or}}\hspace{0.25in} t = – \frac{{19}}{3}\end{align*}\]

\[\begin{align*}1 – 3t & = – 20\hspace{0.25in}{\mbox{or}}\hspace{0.25in}1 – 3t = 20\\ – 3t & = – 21\hspace{0.25in}{\mbox{or}}\hspace{0.25in} – 3t = 19\\ t & = 7\hspace{0.25in}{\mbox{or}}\hspace{0.25in} t = – \frac{{19}}{3}\end{align*}\]

The two solutions to this equation are \(t = – \frac{{19}}{3}\) and \(t = 7\).

c

\(\left| {5y – 8} \right| = 1\)

Again, not much more to this one.

\[\begin{align*}5y – 8 & = – 1 & \hspace{0.25in}& {\mbox{or}} & \hspace{0.25in}5y – 8 & = 1\\ 5y & = 7 & \hspace{0.25in}& {\mbox{or}}& \hspace{0.25in} 5y & = 9\\ y & = \frac{7}{5}& \hspace{0.25in}& {\mbox{or}}& \hspace{0.25in} y & = \frac{9}{5}\end{align*}\]

\[\begin{align*}5y – 8 & = – 1 & \hspace{0.25in}& {\mbox{or}} & \hspace{0.25in}5y – 8 & = 1\\ 5y & = 7 & \hspace{0.25in}& {\mbox{or}}& \hspace{0.25in} 5y & = 9\\ y & = \frac{7}{5}& \hspace{0.25in}& {\mbox{or}}& \hspace{0.25in} y & = \frac{9}{5}\end{align*}\]

In this case the two solutions are \(y = \frac{7}{5}\) and \(y = \frac{9}{5}\).

\(\left| {2x – 5} \right| = 9\)\(\left| {1 – 3t} \right| = 20\)\(\left| {5y – 8} \right| = 1\)

Solve each of the following.

Now, let’s take a look at how to deal with equations for which \(b\) is zero or negative. We’ll do this with an example.

Example 2

Solve each of the following.

  1. \(\left| {10x – 3} \right| = 0\)
  2. \(\left| {5x + 9} \right| = – 3\)

 

a

\(\left| {10x – 3} \right| = 0\)

Let’s approach this one from a geometric standpoint. This is saying that the quantity in the absolute value bars has a distance of zero from the origin. There is only one number that has the property and that is zero itself. So, we must have,

\[10x – 3 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{3}{{10}}\]

\[10x – 3 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{3}{{10}}\]

In this case we get a single solution.

b

\(\left| {5x + 9} \right| = – 3\)

Now, in this case let’s recall that we noted at the start of this section that \(\left| p \right| \ge 0\). In other words, we can’t get a negative value out of the absolute value. That is exactly what this equation is saying however. Since this isn’t possible that means there is no solution to this equation.

\(\left| {10x – 3} \right| = 0\)\(\left| {5x + 9} \right| = – 3\)

Solve each of the following.

So, summarizing we can see that if \(b\) is zero then we can just drop the absolute value bars and solve the equation. Likewise, if \(b\) is negative then there will be no solution to the equation.

To this point we’ve only looked at equations that involve an absolute value being equal to a number, but there is no reason to think that there has to only be a number on the other side of the equal sign. Likewise, there is no reason to think that we can only have one absolute value in the problem. So, we need to take a look at a couple of these kinds of equations.

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Example 3

Solve each of the following.

  1. \(\left| {x – 2} \right| = 3x + 1\)
  2. \(\left| {4x + 3} \right| = 3 – x\)
  3. \(\left| {2x – 1} \right| = \left| {4x + 9} \right|\)

 

At first glance the formula we used above will do us no good here. It requires the right side of the equation to be a positive number. It turns out that we can still use it here, but we’re going to have to be careful with the answers as using this formula will, on occasion introduce an incorrect answer. So, while we can use the formula we’ll need to make sure we check our solutions to see if they really work.

a

\(\left| {x – 2} \right| = 3x + 1\)

So, we’ll start off using the formula above as we have in the previous problems and solving the two linear equations.

\[\begin{align*}x – 2 & = – \left( {3x + 1} \right) = – 3x – 1 & \hspace{0.25in} &{\mbox{or}}& \hspace{0.25in}x – 2 & = 3x + 1\\ 4x & = 1&\hspace{0.25in}&{\mbox{or}} &\hspace{0.25in} – 2x & = 3\\ x & = \frac{1}{4}&\hspace{0.25in}&{\mbox{or}}&\hspace{0.25in} x &= – \frac{3}{2}\end{align*}\]

\[\begin{align*}x – 2 & = – \left( {3x + 1} \right) = – 3x – 1 & \hspace{0.25in} &{\mbox{or}}& \hspace{0.25in}x – 2 & = 3x + 1\\ 4x & = 1&\hspace{0.25in}&{\mbox{or}} &\hspace{0.25in} – 2x & = 3\\ x & = \frac{1}{4}&\hspace{0.25in}&{\mbox{or}}&\hspace{0.25in} x &= – \frac{3}{2}\end{align*}\]

Okay, we’ve got two potential answers here. There is a problem with the second one however. If we plug this one into the equation we get,

\[\begin{align*}\left| { – \frac{3}{2} – 2} \right| & \mathop = \limits^? 3\left( { – \frac{3}{2}} \right) + 1\\ \left| { – \frac{7}{2}} \right|& \mathop = \limits^? – \frac{7}{2}\\ \frac{7}{2} & \ne – \frac{7}{2}\hspace{0.25in}{\mbox{NOT OK}}\end{align*}\]

\[\begin{align*}\left| { – \frac{3}{2} – 2} \right| & \mathop = \limits^? 3\left( { – \frac{3}{2}} \right) + 1\\ \left| { – \frac{7}{2}} \right|& \mathop = \limits^? – \frac{7}{2}\\ \frac{7}{2} & \ne – \frac{7}{2}\hspace{0.25in}{\mbox{NOT OK}}\end{align*}\]

We get the same number on each side but with opposite signs. This will happen on occasion when we solve this kind of equation with absolute values. Note that we really didn’t need to plug the solution into the whole equation here. All we needed to do was check the portion without the absolute value and if it was negative then the potential solution will NOT in fact be a solution and if it’s positive or zero it will be solution.

We’ll leave it to you to verify that the first potential solution does in fact work and so there is a single solution to this equation : \[x = \frac{1}{4}\] and notice that this is less than 2 (as our assumption required) and so is a solution to the equation with the absolute value in it.

So, all together there is a single solution to this equation : \[x = \frac{1}{4}\].

b

\(\left| {4x + 3} \right| = 3 – x\)

This one will work in pretty much the same way so we won’t put in quite as much explanation.

\[\begin{align*}4x + 3 & = – \left( {3 – x} \right) = – 3 + x & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}4x + 3 & = 3 – x\\ 3x & = – 6 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} 5x & = 0\\ x & = – 2 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} x & = 0\end{align*}\]

\[\begin{align*}4x + 3 & = – \left( {3 – x} \right) = – 3 + x & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}4x + 3 & = 3 – x\\ 3x & = – 6 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} 5x & = 0\\ x & = – 2 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} x & = 0\end{align*}\]

Now, before we check each of these we should give a quick warning. Do not make the assumption that because the first potential solution is negative it won’t be a solution. We only exclude a potential solution if it makes the portion without absolute value bars negative. In this case both potential solutions will make the portion without absolute value bars positive and so both are in fact solutions.

So in this case, unlike the first example, we get two solutions : \(x = – 2\) and \(x = 0\).

c

\(\left| {2x – 1} \right| = \left| {4x + 9} \right|\)

This case looks very different from any of the previous problems we’ve worked to this point and in this case the formula we’ve been using doesn’t really work at all. However, if we think about this a little we can see that we’ll still do something similar here to get a solution.

Both sides of the equation contain absolute values and so the only way the two sides are equal will be if the two quantities inside the absolute value bars are equal or equal but with opposite signs. Or in other words, we must have,

\[\begin{align*}2x – 1 & = – \left( {4x + 9} \right) = – 4x – 9 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}2x – 1 & = 4x + 9\\ 6x & = – 8 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} – 2x & = 10\\ x & = – \frac{8}{6} = – \frac{4}{3} & \hspace{0.25in}& {\mbox{or}}& \hspace{0.25in} x & = – 5\end{align*}\]

\[\begin{align*}2x – 1 & = – \left( {4x + 9} \right) = – 4x – 9 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}2x – 1 & = 4x + 9\\ 6x & = – 8 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} – 2x & = 10\\ x & = – \frac{8}{6} = – \frac{4}{3} & \hspace{0.25in}& {\mbox{or}}& \hspace{0.25in} x & = – 5\end{align*}\]

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Now, we won’t need to verify our solutions here as we did in the previous two parts of this problem. Both with be solutions provided we solved the two equations correctly. However, it will probably be a good idea to verify them anyway just to show that the solution technique we used here really did work properly.

Let’s first check \(x = – \frac{4}{3}\).

\[\begin{align*}\left| {2\left( { – \frac{4}{3}} \right) – 1} \right| & \mathop = \limits^? \left| {4\left( { – \frac{4}{3}} \right) + 9} \right|\\ \left| { – \frac{{11}}{3}} \right| & \mathop = \limits^? \left| {\frac{{11}}{3}} \right|\\ \frac{{11}}{3} & = \frac{{11}}{3}\hspace{0.25in}{\mbox{OK}}\end{align*}\]

\[\begin{align*}\left| {2\left( { – \frac{4}{3}} \right) – 1} \right| & \mathop = \limits^? \left| {4\left( { – \frac{4}{3}} \right) + 9} \right|\\ \left| { – \frac{{11}}{3}} \right| & \mathop = \limits^? \left| {\frac{{11}}{3}} \right|\\ \frac{{11}}{3} & = \frac{{11}}{3}\hspace{0.25in}{\mbox{OK}}\end{align*}\]

In the case the quantities inside the absolute value were the same number but opposite signs. However, upon taking the absolute value we got the same number and so \(x = – \frac{4}{3}\) is a solution. Now, let’s check \(x = – 5\).

\[\begin{align*}\left| {2\left( { – 5} \right) – 1} \right| & \mathop = \limits^? \left| {4\left( { – 5} \right) + 9} \right|\\ \left| { – 11} \right| & \mathop = \limits^? \left| { – 11} \right|\\ 11 & = 11\hspace{0.25in}{\mbox{OK}}\end{align*}\]

\[\begin{align*}\left| {2\left( { – 5} \right) – 1} \right| & \mathop = \limits^? \left| {4\left( { – 5} \right) + 9} \right|\\ \left| { – 11} \right| & \mathop = \limits^? \left| { – 11} \right|\\ 11 & = 11\hspace{0.25in}{\mbox{OK}}\end{align*}\]

In the case we got the same value inside the absolute value bars.

So, as suggested above both answers did in fact work and both are solutions to the equation.

\(\left| {x – 2} \right| = 3x + 1\)\(\left| {4x + 3} \right| = 3 – x\)\(\left| {2x – 1} \right| = \left| {4x + 9} \right|\)

Solve each of the following.

So, as we’ve seen in the previous set of examples we need to be a little careful if there are variables on both sides of the equal sign. If one side does not contain an absolute value then we need to look at the two potential answers and make sure that each is in fact a solution.


Math Shorts Episode 10 – Absolute Value


This Math Shorts video defines the term Absolute Value and provides real world examples.
This video was made for the PBS LearningMedia library, thanks to a generous grant from the Corporation for Public Broadcasting. It was produced in collaboration with Utah Education Network.
Common Core State Standards:
CCSS.Math.Content.6.NS.C.7c
7. Understand ordering and absolute value of rational numbers.
c. Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a realworld situation. For example, for an account balance of 30 dollars, write |30| = 30 to describe the size of the debt in dollars.

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Math Shorts Episode 10 - Absolute Value

How to Get an Absolute Value in Excel – Two Examples of its Use


In this video we show how to get the absolute value in Excel using the ABS function and 2 examples of why you may want to do this.
The absolute value in simple terms is the distance of the number from 0. But why would you need this?
In Excel the most common use is to convert a negative value to a positive one within a formula. This video shows 2 examples where this might be useful.
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How to Get an Absolute Value in Excel - Two Examples of its Use

What is Absolute Value? (Concept) | Don’t Memorise


Can you think of one RealLife Application of Absolute Value?
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0:00 Absolute value of a number
0:38 Reallife applications of absolute value
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AbsoluteValue

What is Absolute Value? (Concept) | Don't Memorise

How To Solve Absolute Value Equations


This math video tutorial explains how to solve absolute value equations with variables on both sides. It contains plenty of examples and practice problems.
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How To Solve Absolute Value Equations

Complex Absolute Value Inequality Example 1


Complex Absolute Value Inequality Example 1

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ขอบคุณที่รับชมกระทู้ครับ absolute value

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