What are the solutions to the equation solvefor x,x^2-6xy+y^2=0 ?

The solutions to the equation solvefor x,x^2-6xy+y^2=0 are x=(3+2sqrt(2))y,x=(3-2sqrt(2))y

Find the zeros of solvefor x,x^2-6xy+y^2=0

The zeros of solvefor x,x^2-6xy+y^2=0 are x=(3+2sqrt(2))y,x=(3-2sqrt(2))y

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solve for

x, x^{2} - 6 x y + y^{2} = 0

x = (3 + 22) y, x = (3 - 22) y

Solution steps

x^{2} - 6 x y + y^{2} = 0

Write in the standard form

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

x^{2} - 6 y x + y^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 y ) \pm ( - 6 y ) ^{2} - 4 \cdot 1 \cdot y ^{2}}{2 \cdot 1}

Simplify

(- 6 y)^{2} - 4 \cdot 1 \cdot y^{2} : 42 y

x_{1, 2} = \frac{- ( - 6 y ) \pm 4 2 y}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 y ) + 4 2 y}{2 \cdot 1}, x_{2} = \frac{- ( - 6 y ) - 4 2 y}{2 \cdot 1}

x = \frac{- ( - 6 y ) + 4 2 y}{2 \cdot 1} : (3 + 22) y

x = \frac{- ( - 6 y ) - 4 2 y}{2 \cdot 1} : (3 - 22) y

The solutions to the quadratic equation are:

x = (3 + 22) y, x = (3 - 22) y

x, 2 p x^{2} + 3 x - 2 - 5 p = 0

25 (x + 1) = \frac{7}{2} x^{2}

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

(y - 4) (y - 9) = 0

4 b^{2} = 4 b + 48

What are the solutions to the equation solvefor x,x^2-6xy+y^2=0 ?
The solutions to the equation solvefor x,x^2-6xy+y^2=0 are x=(3+2sqrt(2))y,x=(3-2sqrt(2))y
Find the zeros of solvefor x,x^2-6xy+y^2=0
The zeros of solvefor x,x^2-6xy+y^2=0 are x=(3+2sqrt(2))y,x=(3-2sqrt(2))y