Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Some of the LaTeX for the equations was hard to read. Added a bit more annotation also.)
Line 9: Line 9:
 
First, square both sides, and isolate the absolute value.
 
First, square both sides, and isolate the absolute value.
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
5|x|+8&=x^2-16,\\
+
5|x|+8&=x^2-16\\
−
5|x|&=x^2-24,\\
+
5|x|&=x^2-24\\
−
|x|&=\frac{x^2-24}{5}.
+
|x|&=\frac{x^2-24}{5}. \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
Solve for the absolute value and factor.
 
Solve for the absolute value and factor.
−
[i] Case 1: <math>x=\frac{x^2-24}{5}</math> [/i]
+
 
 +
Case 1: <math>x=\frac{x^2-24}{5}</math>
 +
 
 
Multiplying both sides by <math>5</math> gives us  
 
Multiplying both sides by <math>5</math> gives us  
 
<cmath> 5x=x^2-24.</cmath>
 
<cmath> 5x=x^2-24.</cmath>
 
Rearranging and factoring, we have
 
Rearranging and factoring, we have
−
<math>\begin{align}
+
<cmath>\begin{align*}
 
x^2-5x-24 &=0, \\
 
x^2-5x-24 &=0, \\
−
(x-8)(x-3) &= 0.
+
(x-8)(x-3) &= 0.\\
−
\end{align*}</math>
+
\end{align*}</cmath>
 +
 
 +
Case 2: <math>x=\frac{-x^2+24}{5}</math>
  
−
[i] Case 2: <math>x=\frac{-x^2+24}{5}</math> [/i]
 
 
As above, we multiply both sides by <math>5</math> to find
 
As above, we multiply both sides by <math>5</math> to find
 
<cmath> 5x=-x^2+24.</cmath>
 
<cmath> 5x=-x^2+24.</cmath>
 
Rearranging and factoring gives us
 
Rearranging and factoring gives us
−
<math>\begin{align}
+
<cmath>\begin{align*}
 
x^2+5x-24 &=0, \\
 
x^2+5x-24 &=0, \\
−
(x+8)(x-3) &= 0.
+
(x+8)(x-3) &= 0. \\
−
\end{align*}</math>
+
\end{align*}</cmath>
  
 
Combining these cases, we have <math>x= -8, -3, 3, 8</math>. Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for <math>x</math> 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.
 
Combining these cases, we have <math>x= -8, -3, 3, 8</math>. Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for <math>x</math> 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 <math>|x|=|3|</math>, we have
 
Trying <math>|x|=|3|</math>, we have
−
<math>\begin{align*}
+
<cmath>\begin{align*}
 
\sqrt{5|3|+8}&=\sqrt{3^2-16}, \\
 
\sqrt{5|3|+8}&=\sqrt{3^2-16}, \\
 
\sqrt{15+8}&=\sqrt{9-16}, \\
 
\sqrt{15+8}&=\sqrt{9-16}, \\
−
\sqrt{23}\neq\sqrt{-7}.
+
\sqrt{23}\neq\sqrt{-7}.\\
−
\end{align*}</math>
+
\end{align*}</cmath>
 
Therefore, <math>x = 3</math> and <math> x= -3</math> are extraneous.
 
Therefore, <math>x = 3</math> and <math> x= -3</math> are extraneous.
  
 
Checking <math>|x|=|8|</math>, we have
 
Checking <math>|x|=|8|</math>, we have
−
<math>\begin{align*}
+
<cmath>\begin{align*}
 
\sqrt{5|8|+8}&=\sqrt{8^2-16}, \\
 
\sqrt{5|8|+8}&=\sqrt{8^2-16}, \\
 
\sqrt{40+8}&=\sqrt{64-16}, \\
 
\sqrt{40+8}&=\sqrt{64-16}, \\
−
\sqrt{48}&=\sqrt{48}.
+
\sqrt{48}&=\sqrt{48}.\\
−
\end{align*}</math>
+
\end{align*}</cmath>
  
 
The roots of our original equation are <math>-8</math> and <math>8</math> and product is <math>-8 \times 8 = \boxed{\textbf{(A)} -64}</math>.
 
The roots of our original equation are <math>-8</math> and <math>8</math> and product is <math>-8 \times 8 = \boxed{\textbf{(A)} -64}</math>.

Revision as of 20:51, 20 November 2020

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}\neq\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, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_19&oldid=138042"