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

The solutions to the equation solvefor y,x=y^2-4y+1 are y=2+sqrt(x+3),y=2-sqrt(x+3)

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

The zeros of solvefor y,x=y^2-4y+1 are y=2+sqrt(x+3),y=2-sqrt(x+3)

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solve for

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

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

Solution steps

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

Switch sides

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

Move

x

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x, x^{2} + y^{2} + 5 = 0

x, x^{2} + y^{2} - 6 x + 2 y + 10 = 0

- 2 x^{2} + 9 = 0

4 x^{2} - 6 = 0

2 x^{2} + 2 x = 5

What are the solutions to the equation solvefor y,x=y^2-4y+1 ?
The solutions to the equation solvefor y,x=y^2-4y+1 are y=2+sqrt(x+3),y=2-sqrt(x+3)
Find the zeros of solvefor y,x=y^2-4y+1
The zeros of solvefor y,x=y^2-4y+1 are y=2+sqrt(x+3),y=2-sqrt(x+3)