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

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

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

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

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solve for

x, 2 (x + y) = 1 + (x - y)^{2}

x = y + 1 + 2 y, x = y + 1 - 2 y

Solution steps

2 (x + y) = 1 + (x - y)^{2}

Expand

2 (x + y) : 2 x + 2 y

Expand

1 + (x - y)^{2} : 1 + x^{2} - 2 x y + y^{2}

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

Switch sides

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

Move

2 y

to the left side

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

Move

2 x

to the left side

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

Write in the standard form

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

x^{2} - (2 y + 2) x + 1 + y^{2} - 2 y = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = y + 1 + 2 y, x = y + 1 - 2 y

2 x^{2} + 9 - 10 = 0

- 5 x^{2} - 40 x + 45 = 0

2 x^{2} - 8 x + 4

- \frac{1.85}{20.25} x^{2} + \frac{1.85}{20.25} x + 1.82716049

x, x^{2} - 4 x y + 5 y^{2} - 4 y + 4 = 0

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