Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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\textbf{(E) }12 \qquad | \textbf{(E) }12 \qquad | ||
</math> | </math> | ||
− | + | ==Solutions== | |
− | == Solution 1== | + | === Solution 1=== |
In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The <math>x^2</math> terms cancel nicely. <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24</math> | In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The <math>x^2</math> terms cancel nicely. <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24</math> | ||
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Solution by PancakeMonster2004, explanations added by a1b2. | Solution by PancakeMonster2004, explanations added by a1b2. | ||
− | ==Solution 2 (bad)== | + | ===Solution 2 (bad)=== |
Let <math>u=\sqrt{49-x^2}</math>, and let <math>v=\sqrt{25-x^2}</math>. Then <math>v=\sqrt{u^2-24}</math>. Substituting, we get <math>u-\sqrt{u^2-24}=3</math>. Rearranging, we get <math>u-3=\sqrt{u^2-24}</math>. Squaring both sides and solving, we get <math>u=\frac{11}{2}</math> and <math>v=\frac{11}{2}-3=\frac{5}{2}</math>. Adding, we get that the answer is <math>\boxed{(A) 8}</math> | Let <math>u=\sqrt{49-x^2}</math>, and let <math>v=\sqrt{25-x^2}</math>. Then <math>v=\sqrt{u^2-24}</math>. Substituting, we get <math>u-\sqrt{u^2-24}=3</math>. Rearranging, we get <math>u-3=\sqrt{u^2-24}</math>. Squaring both sides and solving, we get <math>u=\frac{11}{2}</math> and <math>v=\frac{11}{2}-3=\frac{5}{2}</math>. Adding, we get that the answer is <math>\boxed{(A) 8}</math> | ||
Revision as of 19:58, 9 February 2018
Suppose that real number satisfies . What is the value of ?
Contents
[hide]Solutions
Solution 1
In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The terms cancel nicely.
Given that = 3,
Solution by PancakeMonster2004, explanations added by a1b2.
Solution 2 (bad)
Let , and let . Then . Substituting, we get . Rearranging, we get . Squaring both sides and solving, we get and . Adding, we get that the answer is
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.