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

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

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

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

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solve for

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

x = 2 y + 3 y^{2} + f, x = 2 y - 3 y^{2} + f

Solution steps

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

Switch sides

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

Move

f

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 4 y)^{2} - 4 \cdot 1 \cdot (y^{2} - f) : 2 3 y^{2} + f

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 y + 3 y^{2} + f, x = 2 y - 3 y^{2} + f

f (x) = x^{\frac{2}{5}} (3 x + 7)

x^{3} - 2 x^{2} - 15 x + 10 [- 2.1]

f (x) = x + \frac{9}{x} + 2

f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 30

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

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