2011 AMC 10B Problems/Problem 19

Problem

What is the product of all the roots of the equation $\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.$

$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$

Solution 1

First, square both sides, and isolate the absolute value. $\begin{align*} 5|x|+8&=x^2-16\\ 5|x|&=x^2-24\\ |x|&=\frac{x^2-24}{5}. \\ \end{align*}$ Solve for the absolute value and factor.

Case 1: $x=\frac{x^2-24}{5}$

Multiplying both sides by $5$ gives us $5x=x^2-24.$ Rearranging and factoring, we have $\begin{align*} x^2-5x-24 &=0, \\ (x-8)(x+3) &= 0.\\ \end{align*}$

Case 2: $x=\frac{-x^2+24}{5}$

As above, we multiply both sides by $5$ to find $5x=-x^2+24.$ Rearranging and factoring gives us $\begin{align*} x^2+5x-24 &=0, \\ (x+8)(x-3) &= 0. \\ \end{align*}$

Combining these cases, we have $x= -8, -3, 3, 8$ . Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for $x$ back into the original equation to ensure it works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways. Trying $|x|=|3|$ , we have $\begin{align*} \sqrt{5|3|+8}&=\sqrt{3^2-16}, \\ \sqrt{15+8}&=\sqrt{9-16}, \\ \sqrt{23} &\not= \sqrt{-7}.\\ \end{align*}$ Therefore, $x = 3$ and $x= -3$ are extraneous.

Checking $|x|=|8|$ , we have $\begin{align*} \sqrt{5|8|+8}&=\sqrt{8^2-16}, \\ \sqrt{40+8}&=\sqrt{64-16}, \\ \sqrt{48}&=\sqrt{48}.\\ \end{align*}$

The roots of our original equation are $-8$ and $8$ and product is $-8 \times 8 = \boxed{\textbf{(A)} -64}$ .

Solution 2

Square both sides, to get $5|x| + 8 = x^2-16$ . Rearrange to get $x^2 - 5|x| - 24 = 0$ . Seeing that $x^2 = |x|^2$ , substitute to get $|x|^2 - 5|x| - 24 = 0$ . We see that this is a quadratic in $|x|$ . Factoring, we get $(|x|-8)(|x|+3) = 0$ , so $|x| = \{8,-3\}$ . Since the radicand of the equation can't be negative, the sole solution is $|x| = 8$ . Therefore, $x$ can be $8$ or $-8$ . The product is then $-8 \times 8 = \boxed{\textbf{(A)} -64}$ .

Solution 3

First we note that $x \in (-\infty,-4] \cup [4,\infty]$ . This will help us later with finding extraneous solutions. Next, we have two cases: $\text{IF } x\leq 4:$ $\text{ } \sqrt{-5x+8} = \sqrt{x^2-16} \implies x^2+5x-24=0 \rightarrow x = -8,3$ . We note that $3$ is not in the range of possible $x$ 's and thus is not a solution. $\text{IF } x\geq 4:$ $$ (Error compiling LaTeX. Unknown error_msg)\text{ } \sqrt{5x+8} = \sqrt{x^2-16} \implies x^2-5x-24=0 \rightarrow $x = -3,8$ $. We again not that$ -3$is an extraneous solution.

Thus, we have the two solutions$ (Error compiling LaTeX. Unknown error_msg)-8 $and$ 8 $. Therefore product is$ -8 \cdot 8 = \boxed{\textbf{(A)} -64}$.

-ConfidentKoala4

Video Solution

https://youtu.be/ubGzkmVyAwE

~savannahsolver

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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