What are the solutions to the equation solvefor x,(x+y+2z)^2+(x+y-2z)^2=8(x+y)z ?

The solutions to the equation solvefor x,(x+y+2z)^2+(x+y-2z)^2=8(x+y)z are x=-y+2z

Find the zeros of solvefor x,(x+y+2z)^2+(x+y-2z)^2=8(x+y)z

The zeros of solvefor x,(x+y+2z)^2+(x+y-2z)^2=8(x+y)z are x=-y+2z

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solve for

x, (x + y + 2 z)^{2} + (x + y - 2 z)^{2} = 8 (x + y) z

x = - y + 2 z

Solution steps

(x + y + 2 z)^{2} + (x + y - 2 z)^{2} = 8 (x + y) z

Expand

(x + y + 2 z)^{2} + (x + y - 2 z)^{2} : 2 y^{2} + 4 x y + 2 x^{2} + 8 z^{2}

Expand

8 (x + y) z : 8 x z + 8 yz

2 y^{2} + 4 x y + 2 x^{2} + 8 z^{2} = 8 x z + 8 yz

Move

8 yz

to the left side

2 y^{2} + 4 x y + 2 x^{2} + 8 z^{2} - 8 yz = 8 x z

Move

8 x z

to the left side

2 y^{2} + 4 x y + 2 x^{2} + 8 z^{2} - 8 yz - 8 x z = 0

Write in the standard form

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

2 x^{2} + (4 y - 8 z) x + 2 y^{2} + 8 z^{2} - 8 yz = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 4 y - 8 z ) \pm ( 4 y - 8 z ) ^{2} - 4 \cdot 2 ( 2 y ^{2} + 8 z ^{2} - 8 yz )}{2 \cdot 2}

(4 y - 8 z)^{2} - 4 \cdot 2 (2 y^{2} + 8 z^{2} - 8 yz) = 0

x_{1, 2} = \frac{- ( 4 y - 8 z ) \pm 0}{2 \cdot 2}

x = \frac{- ( 4 y - 8 z )}{2 \cdot 2}

Simplify

\frac{- ( 4 y - 8 z )}{2 \cdot 2} : - y + 2 z

x = - y + 2 z

The solution to the quadratic equation is:

x = - y + 2 z

x^{2} = \frac{2}{3} (x + 1)

(x - 3)^{2} - 25 = 0

4 (a x^{2} - 5 x + 3) = 8 x^{2} + b x + 12

m^{2} + 2 m + 10 = 0

3 x + x^{2} - 1 = 0

What are the solutions to the equation solvefor x,(x+y+2z)^2+(x+y-2z)^2=8(x+y)z ?
The solutions to the equation solvefor x,(x+y+2z)^2+(x+y-2z)^2=8(x+y)z are x=-y+2z
Find the zeros of solvefor x,(x+y+2z)^2+(x+y-2z)^2=8(x+y)z
The zeros of solvefor x,(x+y+2z)^2+(x+y-2z)^2=8(x+y)z are x=-y+2z