2015 AMC 10A Problems/Problem 16

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Problem

If $y+4 = (x-2)^2, x+4 = (y-2)^2$, and $x \neq y$, what is the value of $x^2+y^2$?

$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }\text{30}$

Solution

First simplify the equations

$y+4=(x-2)^2$

$y+4=x^2-4x+4$

$y=x^2-4x$ and the the other equation will become $x=y^2-4y$

Substitute $y$ into $x=y^2-4y$ to get

$x=(x^2-4x)^2-4(x^2-4x)$

$x=(x^2-4x)(x^2-4x-4)$

$x=x(x-4)(x^2-4x-4)$

$1=x^3-8x^2+12x+16$

$0=x^3-8x^2+12x+15$

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