2011 AMC 10B Problems/Problem 19

Revision as of 02:50, 10 June 2011 by Gina (talk | contribs) (Solution)

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

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} (Error compiling LaTeX. Unknown error_msg)

Solve for the absolute value and factor. \[x=\frac{x^2-24}{5} \longrightarrow 5x=x^2-24 \longrightarrow 0=x^2-5x-24 \longrightarrow (x-8)(x+3)=0\\ x=\frac{24-x^2}{5} \longrightarrow 5x=24-x^2 \longrightarrow 0=x^2+5x-24 \longrightarrow (x+8)(x-3)=0\] \[x= -8, -3, 3, 8\]

However, this is not the final answer. Plug it back into the original equation to ensure it still works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways. \[\sqrt{5|3|+8}=\sqrt{3^2-16} \longrightarrow\sqrt{15+8}=\sqrt{9-16} \longrightarrow \sqrt{23}\not=\sqrt{-5-16}\\ \sqrt{5|8|+8}=\sqrt{8^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}\] The roots of this equation are $-8$ and $8$ and product is $-8 \times 8 = \boxed{\textbf{(A)} -64}$

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions