Difference between revisions of "2011 AMC 10B Problems/Problem 19"
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== Solution 2 == | == 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 |x|. Factoring, we get <math>(|x|-8)(|x|+3) = 0</math>, so <math>|x| = \{8,-3\}</math>. Since | + | 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 |x|. 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, the x can be <math>8</math> or <math>-8</math>. The product is then <math>-8 \times 8 = \boxed{\textbf{(A)} -64}</math>. |
== See Also== | == See Also== |
Revision as of 12:24, 14 August 2017
Contents
Problem
What is the product of all the roots of the equation
Solution 1
First, square both sides, and isolate the absolute value. Solve for the absolute value and factor.
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. The roots of this equation are and and product is
Solution 2
Square both sides, to get . Rearrange to get . Seeing that , substitute to get . We see that this is a quadratic in |x|. Factoring, we get , so . Since the radicand of the equation can't be negative, the sole solution is . Therefore, the x can be or . The product is then .
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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