2018 AMC 10A Problems/Problem 10

Suppose that real number $x$ satisfies $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ . What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$ ?

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

Solution 1

$(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24$

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

Solution by PancakeMonster2004.

Solution 2

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 $\boxed{(A) 8}$

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2018 AMC 10A Problems/Problem 10

Solution 1

Solution 2

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