Difference between revisions of "2011 AMC 10B Problems/Problem 19"
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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\\ | <cmath>x=\frac{x^2-24}{5} \longrightarrow 5x=x^2-24 \longrightarrow 0=x^2-5x-24 \longrightarrow (x-8)(x+3)=0\\ | ||
Revision as of 13:50, 7 September 2015
Problem
What is the product of all the roots of the equation ![\[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]](http://latex.artofproblemsolving.com/4/d/c/4dc3a78877196f453f6485a158c37ac0179def64.png)
Solution
First, square both sides, and isolate the absolute value.
Solve for the absolute value and factor.
![\[x=\frac{x^2-24}{5} \longrightarrow 5x=x^2-24 \longrightarrow 0=x^2-5x-24 \longrightarrow (x-8)(x+3)=0\\ x=\frac{24-x^2}{5} \longrightarrow 5x=24-x^2 \longrightarrow 0=x^2+5x-24 \longrightarrow (x+8)(x-3)=0\]](http://latex.artofproblemsolving.com/d/6/8/d6830f2f96346ed1cd7f9ae55711126d9ace44af.png)
![\[x= -8, -3, 3, 8\]](http://latex.artofproblemsolving.com/2/b/c/2bc7598b9a1202046784f5096580e2bcea90bc31.png)
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.
![\[\sqrt{5|3|+8}=\sqrt{3^2-16} \longrightarrow\sqrt{15+8}=\sqrt{9-16} \longrightarrow \sqrt{23}\not=\sqrt{-7}\\ \sqrt{5|8|+8}=\sqrt{8^2-16} \longrightarrow \sqrt{40+8}=\sqrt{64-16} \longrightarrow \sqrt{48}=\sqrt{48}\]](http://latex.artofproblemsolving.com/b/a/6/ba65498c58eb767d5052c3394db363d6d4fe1e6d.png)
The roots of this equation are 
and 
and product is 
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. 
