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

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

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

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

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solve for

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

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

Solution steps

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

Move

4

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 x ) \pm ( - 4 x ) ^{2} - 4 \cdot 1 \cdot ( x ^{2} - 4 )}{2 \cdot 1}

Simplify

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

y_{1, 2} = \frac{- ( - 4 x ) \pm 2 3 x ^{2} + 4}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 4 x ) + 2 3 x ^{2} + 4}{2 \cdot 1}, y_{2} = \frac{- ( - 4 x ) - 2 3 x ^{2} + 4}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

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

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

x^{2} - x + 1

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

x^{2} - 2 x + 2

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