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

(Created page with '== Problem 19 == What is the product of all the roots of the equation <cmath>\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.</cmath> <math> \textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qqu…')
 
(→‎Solution 4)
 
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== Problem 19 ==
+
== 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>
Line 5: Line 5:
 
<math> \textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576</math>
 
<math> \textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576</math>
  
−
== Solution ==
+
== Solution 1 ==
  
 
First, square both sides, and isolate the absolute value.
 
First, square both sides, and isolate the absolute value.
Line 11: Line 11:
 
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}</cmath>
+
|x|&=\frac{x^2-24}{5}. \\
 +
\end{align*}</cmath>
 
Solve for the absolute value and factor.
 
Solve for the absolute value and factor.
−
<cmath>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</cmath>
 
−
<cmath>x= -8, -3, 3, 8</cmath>
 
  
−
However, this is not the final answer. Plug it back into the original equation to ensure it still works.
+
Case 1: <math>x=\frac{x^2-24}{5}</math>
−
<cmath>\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}\\
+
Multiplying both sides by <math>5</math> gives us
−
\sqrt{5|3|+8}=\sqrt{3^2-16} \longrightarrow\sqrt{15+8}=\sqrt{9-16} \longrightarrow \sqrt{23}\not=\sqrt{-5-16}\\
+
<cmath> 5x=x^2-24.</cmath>
−
\sqrt{5|8|+8}=\sqrt{8^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}</cmath>
+
Rearranging and factoring, we have
−
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>
+
<cmath>\begin{align*}
 +
x^2-5x-24 &=0, \\
 +
(x-8)(x+3) &= 0.\\
 +
\end{align*}</cmath>
 +
 
 +
Case 2: <math>x=\frac{-x^2+24}{5}</math>
 +
 
 +
As above, we multiply both sides by <math>5</math> to find
 +
<cmath> 5x=-x^2+24.</cmath>
 +
Rearranging and factoring gives us
 +
<cmath>\begin{align*}
 +
x^2+5x-24 &=0, \\
 +
(x+8)(x-3) &= 0. \\
 +
\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.
 +
Trying <math>|x|=|3|</math>, we have
 +
<cmath>\begin{align*}
 +
\sqrt{5|3|+8}&=\sqrt{3^2-16}, \\
 +
\sqrt{15+8}&=\sqrt{9-16}, \\
 +
\sqrt{23} &\not= \sqrt{-7}.\\
 +
\end{align*}</cmath>
 +
Therefore, <math>x = 3</math> and <math> x= -3</math> are extraneous.
 +
 
 +
Checking <math>|x|=|8|</math>, we have
 +
<cmath>\begin{align*}
 +
\sqrt{5|8|+8}&=\sqrt{8^2-16}, \\
 +
\sqrt{40+8}&=\sqrt{64-16}, \\
 +
\sqrt{48}&=\sqrt{48}.\\
 +
\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>.
 +
 
 +
== Solution 2 ==
 +
 
 +
Square both sides, to get <math>5|x| + 8 = x^2-16</math>. Rearrange to get <math>x^2 - 5|x| - 24 = 0</math>. Seeing that <math>x^2 = |x|^2</math>, substitute to get <math>|x|^2 - 5|x| - 24 = 0</math>. We see that this is a quadratic in <math>|x|</math>. Factoring, we get <math>(|x|-8)(|x|+3) = 0</math>, so <math>|x| = \{8,-3\}</math>. Since the radicand of the equation can't be negative, the sole solution is <math>|x| = 8</math>. Therefore, <math>x</math> can be <math>8</math> or <math>-8</math>. The product is then <math>-8 \times 8 = \boxed{\textbf{(A)} -64}</math>.
 +
 
 +
== Solution 3 ==
 +
First we note that <math>x \in (-\infty,-4] \cup [4,\infty]</math>. This will help us later with finding extraneous solutions.
 +
Next, we have two cases:
 +
<cmath>\text{IF } x\leq 4:</cmath>
 +
<cmath>\text{ } \sqrt{-5x+8} = \sqrt{x^2-16} \implies x^2+5x-24=0 \rightarrow x = -8,3</cmath>.
 +
We note that <math>3</math> is not in the range of possible <math>x</math>'s and thus is not a solution.
 +
 
 +
<cmath>\text{IF } x\geq 4:</cmath>
 +
<cmath>\text{ } \sqrt{5x+8} = \sqrt{x^2-16} \implies x^2-5x-24=0 \rightarrow x = -3,8</cmath>.
 +
We again not that <math>-3</math> is an extraneous solution.
 +
 
 +
Thus, we have the two solutions <math>-8</math> and <math>8</math>. Therefore product is <math>-8 \cdot 8 = \boxed{\textbf{(A)} -64}</math>.
 +
 
 +
-ConfidentKoala4
 +
 
 +
== Solution 4 ==
 +
 
 +
To make this problem easier to comprehend, we can define variable <math>a = |x|</math>, with the condition that <math>a</math> is always nonnegative. Also, since any number squared is always nonnegative, we can define <math>|x|^2 = x^2</math>. Then we can square both sides and substitute:
 +
<cmath>\sqrt{5 | x | + 8} = \sqrt{x^2 - 16} => 5a + 8 = a^2 - 16.</cmath>
 +
Bringing the equation over to one side, we get:
 +
<cmath>a^2 - 5a - 24 = 0.</cmath>
 +
Solving for a by factoring, we get:
 +
<cmath>a^2 - 5a - 24 = 0 = (a - 8)(a + 3) = 0,</cmath>
 +
so <math>a = 8</math> or <math>a = -3</math>. But <math>a</math> must be nonnegative, so the only value that works is <math>8</math>. If <math>a = 8</math>, then <math>|x| = 8</math> and <math>x^2 = 8^2 = 64</math>, so <math>x</math> can equal <math>8 or -8</math>. Multiplying the two, we get <math>8 * -8 = \boxed{\textbf{(A)} -64}</math>.
 +
 
 +
~BeepTheSheep954
 +
 
 +
==Video Solution==
 +
https://youtu.be/ubGzkmVyAwE
 +
 
 +
~savannahsolver
 +
 
 +
 
 +
== See Also==
 +
 
 +
{{AMC10 box|year=2011|ab=B|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Latest revision as of 21:36, 19 July 2024

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.

\[\text{IF } x\geq 4:\] \[\text{ } \sqrt{5x+8} = \sqrt{x^2-16} \implies x^2-5x-24=0 \rightarrow x = -3,8\]. We again not that $-3$ is an extraneous solution.

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

-ConfidentKoala4

Solution 4

To make this problem easier to comprehend, we can define variable $a = |x|$, with the condition that $a$ is always nonnegative. Also, since any number squared is always nonnegative, we can define $|x|^2 = x^2$. Then we can square both sides and substitute: \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16} => 5a + 8 = a^2 - 16.\] Bringing the equation over to one side, we get: \[a^2 - 5a - 24 = 0.\] Solving for a by factoring, we get: \[a^2 - 5a - 24 = 0 = (a - 8)(a + 3) = 0,\] so $a = 8$ or $a = -3$. But $a$ must be nonnegative, so the only value that works is $8$. If $a = 8$, then $|x| = 8$ and $x^2 = 8^2 = 64$, so $x$ can equal $8 or -8$. Multiplying the two, we get $8 * -8 = \boxed{\textbf{(A)} -64}$.

~BeepTheSheep954

Video Solution

https://youtu.be/ubGzkmVyAwE

~savannahsolver


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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