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Difference between revisions of "2018 AMC 10A Problems/Problem 10"

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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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Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2})</math> = 3, <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{(A) 8}</math>
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Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2})</math> = 3, <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}</math>
  
 
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 $x$ satisfies \[\sqrt{49-x^2}-\sqrt{25-x^2}=3\]. What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?

$\textbf{(A) }8 \qquad \textbf{(B) }\sqrt{33}+8\qquad \textbf{(C) }9 \qquad \textbf{(D) }2\sqrt{10}+4 \qquad \textbf{(E) }12 \qquad$

Solutions

Solution 1

In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The $x^2$ terms cancel nicely. $(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24$

Given that $(\sqrt {49-x^2} - \sqrt {25-x^2})$ = 3, $(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}$

Solution by PancakeMonster2004, explanations added by a1b2.

Solution 2 (bad)

Let $u=\sqrt{49-x^2}$, and let $v=\sqrt{25-x^2}$. Then $v=\sqrt{u^2-24}$. Substituting, we get $u-\sqrt{u^2-24}=3$. Rearranging, we get $u-3=\sqrt{u^2-24}$. Squaring both sides and solving, we get $u=\frac{11}{2}$ and $v=\frac{11}{2}-3=\frac{5}{2}$. Adding, we get that the answer is $\boxed{(A) 8}$

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

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