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

Revision as of 18:01, 4 February 2015

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

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

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

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

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

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

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

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

@@ Line 1: / Line 1: @@
-lemme get dat pentakill papa johns
+==Problem==
-i need 5 pizzas
+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>