What are the solutions to the equation solvefor x,x^2-y^2=2xy ?

The solutions to the equation solvefor x,x^2-y^2=2xy are x=(1+sqrt(2))y,x=(1-sqrt(2))y

Find the zeros of solvefor x,x^2-y^2=2xy

The zeros of solvefor x,x^2-y^2=2xy are x=(1+sqrt(2))y,x=(1-sqrt(2))y

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solve for

x, x^{2} - y^{2} = 2 x y

x = (1 + 2) y, x = (1 - 2) y

Solution steps

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

Move

2 x y

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 2 y)^{2} - 4 \cdot 1 \cdot (- y^{2}) : 22 y

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

Separate the solutions

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

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

x = \frac{- ( - 2 y ) - 2 2 y}{2 \cdot 1} : (1 - 2) y

The solutions to the quadratic equation are:

x = (1 + 2) y, x = (1 - 2) y

x, 4 - y = 3 (x^{2} + z^{2})

4 x^{2} + 7 = 43

0.6 x^{2} + 2.3 x - 20 = 0

n^{2} - 2 n - 3 = 0

(1 - x) (4 - x) + 2 = 0

What are the solutions to the equation solvefor x,x^2-y^2=2xy ?
The solutions to the equation solvefor x,x^2-y^2=2xy are x=(1+sqrt(2))y,x=(1-sqrt(2))y
Find the zeros of solvefor x,x^2-y^2=2xy
The zeros of solvefor x,x^2-y^2=2xy are x=(1+sqrt(2))y,x=(1-sqrt(2))y