What are the solutions to the equation solvefor x,x^2-4xy+y^2=0 ?

The solutions to the equation solvefor x,x^2-4xy+y^2=0 are x=(2+sqrt(3))y,x=(2-sqrt(3))y

Find the zeros of solvefor x,x^2-4xy+y^2=0

The zeros of solvefor x,x^2-4xy+y^2=0 are x=(2+sqrt(3))y,x=(2-sqrt(3))y

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solve for

x, x^{2} - 4 x y + y^{2} = 0

x = (2 + 3) y, x = (2 - 3) y

Solution steps

x^{2} - 4 x y + y^{2} = 0

Write in the standard form

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

x^{2} - 4 y x + y^{2} = 0

Solve with the quadratic formula

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

Simplify

(- 4 y)^{2} - 4 \cdot 1 \cdot y^{2} : 23 y

x_{1, 2} = \frac{- ( - 4 y ) \pm 2 3 y}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 y ) + 2 3 y}{2 \cdot 1}, x_{2} = \frac{- ( - 4 y ) - 2 3 y}{2 \cdot 1}

x = \frac{- ( - 4 y ) + 2 3 y}{2 \cdot 1} : (2 + 3) y

x = \frac{- ( - 4 y ) - 2 3 y}{2 \cdot 1} : (2 - 3) y

The solutions to the quadratic equation are:

x = (2 + 3) y, x = (2 - 3) y

w^{2} - 10 w + 21 = 0

12 x^{2} + 192 x = 0

(x + 6)^{2} = - 16

y, x^{2} - y + 1 = x - y + 1

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

What are the solutions to the equation solvefor x,x^2-4xy+y^2=0 ?
The solutions to the equation solvefor x,x^2-4xy+y^2=0 are x=(2+sqrt(3))y,x=(2-sqrt(3))y
Find the zeros of solvefor x,x^2-4xy+y^2=0
The zeros of solvefor x,x^2-4xy+y^2=0 are x=(2+sqrt(3))y,x=(2-sqrt(3))y