2018 AMC 10A Problems/Problem 10

Problem

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

We let $a=\sqrt{49-x^2}+\sqrt{25-x^2}$ ; in other words, we want to find $a$ . We know that $a\cdot3=\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)\cdot\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=\left(\sqrt{49-x^2}\right)^2-\left(\sqrt{25-x^2}\right)^2=\left(49-x^2\right)-\left(25-x^2\right)=24.$ Thus, $a=\boxed{8}$ .

~Technodoggo

Solution 2

Let $a = \sqrt{49-x^2}$ , and $b = \sqrt{25-x^2}$ . Solving for the constants in terms of x, a , and b, we get $a^2 + x^2 = 49$ , and $b^2 + x^2 = 25$ . Subtracting the second equation from the first gives us $a^2 - b^2 = 24$ . Difference of squares gives us $(a+b)(a-b) = 24$ . Since we want to find $a+b = \sqrt{49-x^2}+\sqrt{25-x^2}$ , and we know $a-b = 3$ , we get $3(a+b) = 24$ , so $a+b = \boxed{\textbf{(A) }8}$

~idk12345678

Solution 3

We can substitute $25 - x^2$ for $a$ , thus turning the equation into $\sqrt{a+24} - \sqrt{a} = 3$ . Moving the $\sqrt{a}$ to the other side and squaring gives us $a + 24 = 9 + 6\sqrt{a} + a$ , solving for $a$ gives us 25/4. We substitute this value into the expression they asked us to evaluate giving 8.

~ SAMANTAP

Solution 4

Move $-\sqrt{25-x^2}$ to the right to get $\sqrt{49-x^2} = 3 + \sqrt{25-x^2}$ . Square both sides to get $49-x^2 = 9 + 6\sqrt{25-x^2} + (25-x^2)$ . Simplify to get $15 = 6\sqrt{25-x^2}$ , or $\frac{5}{2} = \sqrt{25-x^2}$ Substitute this back into the original equation tog et that $\sqrt{49-x^2} = \frac{11}{2}$ . The answer is $\boxed{\textbf{(A) }8}$

-Failure.net

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/P-atxiiTw2I

~Education, the Study of Everything

Video Solutions

Video Solution 1

https://youtu.be/ba6w1OhXqOQ?t=1403

~ pi_is_3.14

Video Solution 2

https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go

Video Solution 3

https://youtu.be/ZiZVIMmo260

Video Solution 4

https://youtu.be/5cA87rbzFdw

~savannahsolver

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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