What are the solutions to the equation solvefor y,4x^2-4xy+1-y^2=0 ?

The solutions to the equation solvefor y,4x^2-4xy+1-y^2=0 are y=-2x-sqrt(8x^2+1),y=-2x+sqrt(8x^2+1)

Find the zeros of solvefor y,4x^2-4xy+1-y^2=0

The zeros of solvefor y,4x^2-4xy+1-y^2=0 are y=-2x-sqrt(8x^2+1),y=-2x+sqrt(8x^2+1)

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solve for

y, 4 x^{2} - 4 x y + 1 - y^{2} = 0

y = - 2 x - 8 x^{2} + 1, y = - 2 x + 8 x^{2} + 1

Solution steps

4 x^{2} - 4 x y + 1 - y^{2} = 0

Write in the standard form

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

- y^{2} - 4 x y + 4 x^{2} + 1 = 0

Solve with the quadratic formula

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

Simplify

(- 4 x)^{2} - 4 (- 1) (4 x^{2} + 1) : 2 8 x^{2} + 1

y_{1, 2} = \frac{- ( - 4 x ) \pm 2 8 x ^{2} + 1}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- ( - 4 x ) + 2 8 x ^{2} + 1}{2 ( - 1 )}, y_{2} = \frac{- ( - 4 x ) - 2 8 x ^{2} + 1}{2 ( - 1 )}

y = \frac{- ( - 4 x ) + 2 8 x ^{2} + 1}{2 ( - 1 )} : - 2 x - 8 x^{2} + 1

y = \frac{- ( - 4 x ) - 2 8 x ^{2} + 1}{2 ( - 1 )} : - 2 x + 8 x^{2} + 1

The solutions to the quadratic equation are:

y = - 2 x - 8 x^{2} + 1, y = - 2 x + 8 x^{2} + 1

x^{2} - 5 = 95

x^{2} + 20 x + 80 = 0

2 x^{2} - 3 x - 44 = 0

a^{2} + 4 a + 20 = 0

5 x^{2} + 21 x + 18 = 0

What are the solutions to the equation solvefor y,4x^2-4xy+1-y^2=0 ?
The solutions to the equation solvefor y,4x^2-4xy+1-y^2=0 are y=-2x-sqrt(8x^2+1),y=-2x+sqrt(8x^2+1)
Find the zeros of solvefor y,4x^2-4xy+1-y^2=0
The zeros of solvefor y,4x^2-4xy+1-y^2=0 are y=-2x-sqrt(8x^2+1),y=-2x+sqrt(8x^2+1)