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2011 AMC 10B Problems/Problem 19

Revision as of 11:24, 14 August 2017 by Progamexd (talk | contribs) (→‎Solution 2)

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. \[x=\frac{x^2-24}{5} \longrightarrow 5x=x^2-24 \longrightarrow 0=x^2-5x-24 \longrightarrow (x-8)(x+3)=0\\ x=\frac{24-x^2}{5} \longrightarrow 5x=24-x^2 \longrightarrow 0=x^2+5x-24 \longrightarrow (x+8)(x-3)=0\] \[x= -8, -3, 3, 8\]

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. \[\sqrt{5|3|+8}=\sqrt{3^2-16} \longrightarrow\sqrt{15+8}=\sqrt{9-16} \longrightarrow \sqrt{23}\not=\sqrt{-7}\\ \sqrt{5|8|+8}=\sqrt{8^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}\] The roots of this equation are $-8$ and $8$ and product is $-8 \times 8 = \boxed{\textbf{(A)} -64}$

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, the x can be $8$ or $-8$. The product is then $-8 \times 8 = \boxed{\textbf{(A)} -64}$.

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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