System of equations?

Algebra Level 4

{ y + 4 = ( x 2 ) 2 x + 4 = ( y 2 ) 2 \large{ \begin{cases} y+4 = (x-2)^2 \\ x+ 4 = (y-2)^2 \end{cases} }

Let x x and y y be distinct numbers satisfying the system of equations above.

What is the value of x 2 + y 2 x^2 + y^2 ?


The answer is 15.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Pablo Ruiz
Oct 10, 2016

expanding y + 4 = ( x 2 ) 2 y+4=(x-2)^2 and x + 4 = ( y 2 ) 2 x+4=(y-2)^2

we get y = x 2 4 x y=x^2-4x and x = y 2 4 y x=y^2-4y

x y = ( y 2 ) 2 ( x 2 ) 2 x-y = (y-2)^2 - (x-2)^2

Difference of squares

x y = [ ( y 2 ) ( x 2 ) ] [ ( y 2 ) 2 + ( x 2 ) 2 ] x-y = [(y-2)-(x-2)]*[(y-2)^2+(x-2)^2]

x y = ( y x ) ( y + x 4 ) x-y = (y-x)*(y+x-4)

x y = ( 1 ) ( x y ) ( y + x 4 ) x-y = (-1)*(x-y)*(y+x-4)

( x y ) / [ ( 1 ) ( x y ) ] = ( y + x 4 ) (x-y)/[(-1)*(x-y)]=(y+x-4)

1 = y + x 4 -1=y+x-4

x + y = 3 x+y=3

then

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

x 2 + y 2 = 5 x + 5 y x^2+y^2=5x+5y

x 2 + y 2 = 5 ( x + y ) x^2+y^2=5*(x+y)

x 2 + y 2 = 5 ( 3 ) x^2+y^2=5*(3)

x 2 + y 2 = 15 x^2+y^2=15

Nice one. Did it same way.

ADAMS AYOADE - 4 years, 8 months ago

Excellent. I suppose "me three."

Ian Limarta - 4 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...