SAT Absolute Value

To solve problems with absolute values on the SAT, you need to know:

The definition of absolute value

If $x$ is a real number,

$|x| = \left\{ \begin{array}{lr} x & \quad \text{if}\ \ x > 0,\\ 0 & \quad \text{if}\ \ x=0,\\ -x & \quad \text{if}\ \ x < 0. \end{array} \right.$

We call the quantity inside the absolute value the argument of the absolute value.

When solving equations and inequalities involving absolute value, there are two cases to consider: when the argument of the absolute value is positive, and when it is negative.

For any $a \geq 0$ ,

if $|x|=a$ , then $x=a$ or $x = -a.$

if $|x|\leq a$ , then $-a\leq x \leq a.$

if $|x| \geq a$ , then $x \leq -a$ or $x \geq a.$

the properties of inequality
how to solve inequalities
how to solve simple equations

Examples

The line above has the equation $y=x-4.$ Which of the following is the graph of $y=|x-4|?$

Correct Answer: D

Solution 1:

Tip: Use a calculator.

We use a graphing calculator to graph $y=|x-4|.$ Above, $y=x-4$ is shown in black, and $y=|x-4|$ is shown in green. The graph in choice (D) matches our graph, and hence it is the correct answer.

Solution 2:

The original function, $y=x-4$ , may yield nonnegative or negative outputs. $y=|x-4|$ will not change the original function's nonnegative outputs, but it will make all of its negative outputs positive. On a graph, this means that, when taking the absolute value of a function, all the values of a the original function located below the $x-$ axis (shaded in the graph above) reflect over the $x-$ axis, as shown. The graph in choice (D) is the only one that matches this result.

Solution 3:

Tip: Replace variables with numbers.
In the following table we compute several values of the function $y=|x-4|.$

$\begin{array}{|c|c|c|c|c|c|} \hline x & 2 & 3 & 4 & 5 & 6\\ \hline y=|x-4| & 2 & 1 & 0 & 1 & 2\\ \hline \end{array}$

and in the graph below, we plot these points.

The graph in choice (D) matches our graph, and hence it is the correct answer.

Solution 4:

Tip: Look for a counter-example.
At $x=0, y=|x-4|=|0-4|=|-4|=4.$ In choices (A) and (E), $y=-4,$ so we eliminate them.

At $x=1, y=|x-4|=|1-4|=|-3|=3.$ In choice (B), $y=5$ , and in choice (E), $y=-3.$ We eliminate these choices.

The graphs of choices (C) and (D) are identical when $x \leq 4.$ Let's select a value of $x$ for which they differ, maybe $x=5$ . At $x=5, y=|x-4|=|5-4|=|1|=1.$ In choice (C), $y=-1,$ whereas in choice (D), $y=1.$ We found a counter-example for all choices except for choice (D). It is the correct answer.

Incorrect Choices:

(A)
This is the graph of the function $y=-x-4.$ See Solution 4 for a counter-example.

(B)
This is the graph of the function $y=|x|+4.$ See Solution 4 for a counter-example.

(C)
This is the graph of the function $y=-x+4.$ See Solution 4 for a counter-example.

(E)
This is the graph of the function $y=-|x-4|.$ See Solution 4 for a counter-example.

If $x$ is an integer, how many possible values for $x$ are there that satisfy the inequality $|2-x|<-5$ ?

(A) Zero
(B) Eight
(C) Nine
(D) Eleven
(E) More than eleven

Correct Answer: A

Solution:

By the definition of absolute value, $|2-x|$ is nonnegative. Hence, it cannot be less than -5, and there is no solution.

Incorrect Choices:

(B)
Tip: Read the entire question carefully.
Tip: If you can, verify your choice.
If you solve for all the integers $x$ that satisfy the inequality $|2-x|<\fbox{5},$ you will get $x=$ -2, -1, 0, 1, 2, 3, 4, 5, and 6. If you then miscount, or if you think $0$ is not an integer, you may get this wrong answer. The inequality in the prompt is actually $|2-x|<-5.$

You could verify your result. If $x=1, |2-x| < -5$ will become $|2-1|< -5,$ or $1< -5,$ which is false. Therefore, you should look for a mistake.

(C)
Tip: Read the entire question carefully.
Tip: If you can, verify your choice.
If you solve for all the integers $x$ that satisfy the inequality $|2-x|<\fbox{5},$ you will get $x=$ -2, -1, 0, 1, 2, 3, 4, 5, and 6, or nine values for $x.$ But, the inequality in the prompt is actually $|2-x|<-5.$

You could verify your result. If $x=0, |2-x| < -5$ will become $|2-0| < -5,$ or $0 < -5,$ which is false. You should look for a mistake.

(D)
Tip: Read the entire question carefully.
Tip: If you can, verify your choice.
If you solve for all the integers $x$ that satisfy the inequality $|2-x|\boxed{\leq 5},$ you will get $x=$ -3, -2, -1, 0, 1, 2, 3, 4, 5, and 6, 7, or eleven values for $x.$ But, the inequality in the prompt is actually $|2-x|<-5.$

You could verify your result. If $x=0, |2-x| < -5$ will become $|2-0|< -5,$ or $2< -5,$ which is false, and you should look for a mistake.

(E)
Tip: Read the entire question carefully.
Tip: If you can, verify your choice.
If you pay no attention to the negative sign in the inequality $|2-x|<-5,$ and proceed to solve the two inequalities, $2-x<-5$ and $-(2-x)<-5,$ you will obtain $x>7$ or $x<-3,$ respectively. Since there are many integers that satisfy one or the other inequality, you may conclude this answer is true. But, by definition, $|2-x|$ is nonnegative, and you shouldn't try to solve $|2-x|<-5.$ $x=8$ is counter-example to this result. If $x=8, |2-x|<-5$ becomes $|2-8|<-5.$ Simplifying the left side, we get $|-6|<-5,$ or $6<-5,$ which is false.

If $|x|>|y|,$ which of the following options is always true?

$\begin{array}{r r l} &\text{I.} &|x|+x \geq 0\\ &\text{II.} &\sqrt{x^{2}}<\sqrt{y^{2}}\\ &\text{III.} &x>y\\ \end{array}$

(A) None
(B) I only
(C) II only
(D) I and II only
(E) II and III only

Correct Answer: B

Solution:

Tip: Look for a counter-example.
We analyze each of the options, starting with option I. Let $a$ be a real number, such that $a\geq 0.$

If we let $x=-a$ , then $|-a|+(-a)=a-a=0 \geq 0.$
If we let $x=0$ , then $|0|+0=0 \geq 0.$
If we let $x = a$ , then $|a|+a=a+a=2a \geq 0.$

This option is always true and therefore we can eliminate choice (A).

Now we analyze option II. Here is a counter-example: if $x=2$ and $y=1$ , the given condition, $|x|>|y|$ , is satisfied. Assuming II is true, we have:

$\begin{aligned} \sqrt{x^{2}}&<\sqrt{y^{2}}\\ \sqrt{2^{2}}&<\sqrt{1^{2}}\\ \sqrt{4}&<\sqrt{1}\\ 2&<1\\ \end{aligned}$

But $2>1$ . We have reached a contradiction. Therefore option II is false.

Here is a more abstract approach (you won't be expected to know how to do this on the SAT). If $a$ is a real number, then $a^{2} = (+a)^{2}=(-a)^{2}$ and therefore, both $+a$ and $-a$ are square roots of $a^{2}$ . It follows that

$\begin{array}{l c l l} \sqrt{a^{2}} &=& +a &\quad \text{if}\ a > 0,\\ \sqrt{a^{2}}&=&0 &\quad \text{if}\ a=0, \text{and}\\ \sqrt{a^{2}} &=& -a &\quad \text{if}\ a < 0\\ \end{array}$

or, $\sqrt{a^{2}}=|a|$ .

This means that $|x| = \sqrt{x^{2}}$ and $|y|=\sqrt{y^{2}}$ . And since $|x|>|y|$ , it follows that $\sqrt{x^{2}}>\sqrt{y^{2}}$ , not $\sqrt{x^{2}}<\sqrt{y^{2}}$ .

Therefore, option II is false.

Last, we analyze option III. We find a counter-example: if $x=-2$ and $y=1$ , then the given condition, $|x|>|y|$ , is satisfied. If III were true, then $-2>1$ , which is false.

Answer choices (C), (D), and (E) contain option II or option III, or both, so we eliminate them.

The only choice we haven't eliminated is choice (B), and it is the correct answer.

Incorrect Choices:

(A), (C), (D), and (E)
See the solution above for how to eliminate these choices.

Review

If you thought these examples difficult and you need to review the material, these links will help:

SAT Tips for Absolute Value

For any $a \geq 0$ , if $|x|=a$ , then $x=a$ or $x = -a.$

For any $a \geq 0$ , if $|x|\leq a$ , then $-a\leq x \leq a.$

For any $a \geq 0$ , if $|x| \geq a$ , then $x \leq -a$ or $x \geq a.$

Know the Properties of Inequality.

SAT General Tips

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