Absolute value

To check whether the identity is always true, we use the definitions of absolute value and the principal square root from Number and Algebra Foundations.

1. Absolute value is defined by cases: $\{\begin{array}{ll}x& \text{if }x\ge 0\\ -x& \text{if }x<0\end{array}$

2. The notation $\sqrt{k}$ means the principal square root, which is the non-negative number whose square equals $k$.

Now consider any value of $x$ that is zero or positive.

• Left side: if $x\ge 0$, then by definition $|x|=x$.
• Right side: when $x$ is non-negative, $\sqrt{x^2}=x$.

For instance, if $x=5$: $\begin{array}{rl}|5|& =5\\ \sqrt{5^2}& =\sqrt{25}=5\end{array}$ So the two sides match for all $x\ge 0$.

Next, test what happens for negative inputs.

• Left side: if $x<0$, then $|x|=-x$ (which is positive).
• Right side: although $x$ is negative, $x^2$ is positive, and $\sqrt{x^2}$ must be the non-negative root, which equals $-x$.

For example, if $x=-5$: $\begin{array}{rl}|-5|& =5\\ \sqrt{(-5{)}^2}& =\sqrt{25}=5\end{array}$ Both sides give the same positive value, so the identity also holds when $x<0$.

Because the equality worked in the non-negative case and the negative case (and it clearly works at $x=0$ too), any option claiming it fails or only works on one side of zero must be wrong.

• Option B is wrong because we have shown it works in every sign case.
• Option C is wrong because it also works for $x<0$.
• Option D is wrong because it also works for $x>0$.

Since $|x|$ and $\sqrt{x^2}$ produce the same non-negative output for every real number $x$, the identity is true for all real numbers.

We are checking whether $|x|=\sqrt{x^2}$ is always true for real numbers. A quick way to build confidence is to plug in a positive number, zero, and a negative number and compare both sides.

Test $x=5$:

Left side: $|5|=5$ Right side: $\sqrt{5^2}=\sqrt{25}=5$

Because both sides equal 5, the identity works for a positive number.

Test $x=0$:

Left side: $|0|=0$ Right side: $\sqrt{0^2}=\sqrt{0}=0$

Both sides match, so it also works at zero.

Now test a negative value, such as $x=-3$:

Left side: $|-3|=3$ Right side: $\sqrt{(-3{)}^2}=\sqrt{9}$

Because $\sqrt{\dots }$ means the principal square root (never negative), $\sqrt{9}=3$.

So $|-3|=3$ and $\sqrt{(-3{)}^2}=3$, meaning it works for negative numbers too.

Our checks succeeded for a positive number, for zero, and for a negative number. Therefore, any option saying the identity is false or only true on one side of zero cannot be correct.

Since the equality $|x|=\sqrt{x^2}$ holds for all the types of real numbers we tested (positive, zero, negative), the correct choice is True.

Another way to decide if $|x|=\sqrt{x^2}$ always holds is to graph both sides as functions: $y_1=|x|$ and $y_2=\sqrt{x^2}$. If the graphs coincide for all real $x$, then the identity is valid everywhere.

On a GDC, enter both functions.

1. The graph of $y_1=|x|$ forms a V-shape with its turning point at $(0,0)$ because it measures how far $x$ is from 0.
2. The graph of $y_2=\sqrt{x^2}$ first makes the input non-negative by squaring, then applies the principal square root, which outputs a non-negative value.

When graphed, the two curves overlap completely across the displayed domain.

To support what the graph suggests, check a negative x-value such as $x=-3$:

$\begin{array}{rl}|-3|& =3\\ \sqrt{(-3{)}^2}& =\sqrt{9}=3\end{array}$

The outputs agree for negative inputs, and the same agreement happens for 0 and for positive inputs. This contradicts any claim that the identity only works for part of the real line.

From the overlapping graphs:

• Option B is incorrect because the two graphs match everywhere shown.
• Option C is incorrect because the match also occurs for negative values (for example, $x=-3$), not only for $x\ge 0$.
• Option D is incorrect because the identity is also true for positive values.

Because $y=|x|$ and $y=\sqrt{x^2}$ produce identical graphs for all real $x$, the statement is True.

We need to find the value of the expression $3|x|-|x|$ after replacing $x$ with $-4$. This uses absolute value, which appears in the Number and Algebra Foundations topic.

Recall: the absolute value $|x|$ tells how far $x$ is from 0 on the number line, so it is never negative.

Using Direct Substitution, substitute $x=-4$ everywhere in the expression:

$3|-4|-|-4|$

Now evaluate the absolute value parts. A negative number becomes positive inside absolute value:

$|-4|=4$

Put this back into the expression:

$3(4)-4$

Follow order of operations (multiply before subtracting):

$\begin{array}{rl}3(4)-4& =12-4\\ & =8\end{array}$

Compare our result with the answer choices:

• Option B and Option C cannot be correct because they are negative; with $|x|$ being positive here, the expression gives a positive result.
• Option D would happen if we mistakenly added ($12+4$) instead of subtracting.

Our computed value is $8$, which corresponds to Option A.

A quicker way is to use algebraic simplification before plugging in $x=-4$. We can combine like terms that contain the absolute value $|x|$.

This is part of Number and Algebra Foundations, where we can handle $|x|$ like a single object (similar to a variable such as $a$ or $y$).

Treat $|x|$ as a common factor. Subtract one $|x|$ from three $|x|$:

$\begin{array}{rl}3|x|-|x|& =3|x|-1|x|\\ & =(3-1)|x|\\ & =2|x|\end{array}$

Now evaluate $2|x|$ at $x=-4$. Since $|-4|$ is the distance from 0, it equals 4:

$\begin{array}{rl}2|-4|& =2\times 4\\ & =8\end{array}$

Using the value we found:

• Option B and Option C are negative, but $|-4|$ is positive (4) and multiplying by 2 keeps it positive.
• Option D equals 16, which would come from adding ($3|x|+|x|$) instead of subtracting.

After simplifying to $2|x|$ and then using $x=-4$, the expression evaluates to 8.