Difference between revisions of "2015 AMC 10A Problems/Problem 16"

Revision as of 19:11, 4 February 2015

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

Our equations simplify to (after subtracting 4 from both sides): $y = x^2 - 4x,$ $x = y^2 - 4y.$ Subtract the equations to obtain $y - x = x^2 - y^2 - 4x + 4y$ , so $x^2 - y^2 = 3x - 3y$ . This factors as $(x - y)(x + y) = 3(x - y)$ , and so because $x \neq y$ , we have $x + y = 3$ .

Add the equations to yield $x + y = x^2 + y^2 - 4(x + y)$ . Hence, $x^2 + y^2 = 5(x + y) = 15$ , so our answer is $\boxed{\textbf{(B)}}$ .

@@ Line 12: / Line 12: @@
 Add the equations to yield <math>x + y = x^2 + y^2 - 4(x + y)</math>. Hence, <math>x^2 + y^2 = 5(x + y) = 15</math>, so our answer is <math>\boxed{\textbf{(B)}}</math>.
-==Solution 2==
-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>

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Difference between revisions of "2015 AMC 10A Problems/Problem 16"

Revision as of 19:11, 4 February 2015

Problem

Solution