What are the solutions to the equation solvefor x,2(x^2+y^2)=(x-y)^2+16 ?

The solutions to the equation solvefor x,2(x^2+y^2)=(x-y)^2+16 are x=-y-4,x=-y+4

Find the zeros of solvefor x,2(x^2+y^2)=(x-y)^2+16

The zeros of solvefor x,2(x^2+y^2)=(x-y)^2+16 are x=-y-4,x=-y+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

x, 2 (x^{2} + y^{2}) = (x - y)^{2} + 16

x = - y - 4, x = - y + 4

Solution steps

2 (x^{2} + y^{2}) = (x - y)^{2} + 16

Expand

2 (x^{2} + y^{2}) : 2 x^{2} + 2 y^{2}

Expand

(x - y)^{2} + 16 : x^{2} - 2 x y + y^{2} + 16

2 x^{2} + 2 y^{2} = x^{2} - 2 x y + y^{2} + 16

Switch sides

x^{2} - 2 x y + y^{2} + 16 = 2 x^{2} + 2 y^{2}

Move

2 y^{2}

to the left side

- y^{2} - 2 x y + x^{2} + 16 = 2 x^{2}

Move

2 x^{2}

to the left side

- x^{2} - 2 x y - y^{2} + 16 = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} - 2 y x - y^{2} + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 y ) \pm ( - 2 y ) ^{2} - 4 ( - 1 ) ( - y ^{2} + 16 )}{2 ( - 1 )}

(- 2 y)^{2} - 4 (- 1) (- y^{2} + 16) = 8

x_{1, 2} = \frac{- ( - 2 y ) \pm 8}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 2 y ) + 8}{2 ( - 1 )}, x_{2} = \frac{- ( - 2 y ) - 8}{2 ( - 1 )}

x = \frac{- ( - 2 y ) + 8}{2 ( - 1 )} : - y - 4

x = \frac{- ( - 2 y ) - 8}{2 ( - 1 )} : - y + 4

The solutions to the quadratic equation are:

x = - y - 4, x = - y + 4

5 x^{2} + 3 x - 3 = 3 x + 122

v^{(2)} - 7 v + 6 = 0

x^{2} - 14 x + 10 = 0

3 x^{2} - 196 x + 2376 = 0

(3 x - 2)^{2} = x^{2} + (2 x + 2)^{2}

What are the solutions to the equation solvefor x,2(x^2+y^2)=(x-y)^2+16 ?
The solutions to the equation solvefor x,2(x^2+y^2)=(x-y)^2+16 are x=-y-4,x=-y+4
Find the zeros of solvefor x,2(x^2+y^2)=(x-y)^2+16
The zeros of solvefor x,2(x^2+y^2)=(x-y)^2+16 are x=-y-4,x=-y+4