Difference between revisions of "2011 AMC 10B Problems/Problem 19"

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== Problem 19 ==
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== Problem==
  
 
What is the product of all the roots of the equation <cmath>\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.</cmath>
 
What is the product of all the roots of the equation <cmath>\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.</cmath>
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\sqrt{5|8|+8}=\sqrt{8^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}</cmath>
 
\sqrt{5|8|+8}=\sqrt{8^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}</cmath>
 
The roots of this equation are <math>-8</math> and <math>8</math> and product is <math>-8 \times 8 = \boxed{\textbf{(A)} -64}</math>
 
The roots of this equation are <math>-8</math> and <math>8</math> and product is <math>-8 \times 8 = \boxed{\textbf{(A)} -64}</math>
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== See Also==
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{{AMC10 box|year=2011|ab=B|num-b=18|num-a=20}}

Revision as of 19:50, 4 June 2011

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. \[\sqrt{5|-8|+8}=\sqrt{(-8)^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}\\ \sqrt{5|-3|+8}=\sqrt{(-3)^2-16} \longrightarrow \sqrt{15+8}=\sqrt{9-16} \longrightarrow \sqrt{23}\not=\sqrt{-5}\\ \sqrt{5|3|+8}=\sqrt{3^2-16} \longrightarrow\sqrt{15+8}=\sqrt{9-16} \longrightarrow \sqrt{23}\not=\sqrt{-5-16}\\ \sqrt{5|8|+8}=\sqrt{8^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}\] 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