Difference between revisions of "2015 AMC 10A Problems/Problem 16"
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− | + | ==Problem== | |
− | + | ||
+ | If <math>y+4 = (x-2)^2, x+4 = (y-2)^2</math>, and <math>x \neq y</math>, what is the value of <math>x^2+y^2</math>? | ||
+ | |||
+ | <math> \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }\text{30} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | First simplify the equations | ||
+ | |||
+ | <math>y+4=(x-2)^2</math> | ||
+ | |||
+ | <math>y+4=x^2-4x+4</math> | ||
+ | |||
+ | <math>y=x^2-4x</math> and the the other equation will become <math>x=y^2-4y</math> | ||
+ | |||
+ | Substitute <math>y</math> into <math>x=y^2-4y</math> to get | ||
+ | |||
+ | <math>x=(x^2-4x)^2-4(x^2-4x)</math> | ||
+ | |||
+ | <math>x=(x^2-4x)(x^2-4x-4)</math> | ||
+ | |||
+ | <math>x=x(x-4)(x^2-4x-4)</math> | ||
+ | |||
+ | <math>1=x^3-8x^2+12x+16</math> | ||
+ | |||
+ | <math>0=x^3-8x^2+12x+15</math> |
Revision as of 18:01, 4 February 2015
Problem
If , and , what is the value of ?
Solution
First simplify the equations
and the the other equation will become
Substitute into to get