What is the answer to solvefor y,(sqrt(y+1)-1)^2-2=x ?

The answer to solvefor y,(sqrt(y+1)-1)^2-2=x is y=x+2+2sqrt(x+2),y=x+2-2sqrt(x+2)

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solvefor y,(sqrt(y+1)-1)^2-2=x

solve for

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

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

Solution steps

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

Expand

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

y - 2 y + 1 = x

Remove square roots

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

Solve

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

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

Verify Solutions:

y = x + 2 + 2 x + 2

True

, y = x + 2 - 2 x + 2

True

The solutions are

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

l, 40 = k^{0.5} l^{0.5}

7 = x - 1

(2 - x) (x + 2) = x

x^{2} + 3 + 2 x - 6 = 11

2 x = 17 x - 15

What is the answer to solvefor y,(sqrt(y+1)-1)^2-2=x ?
The answer to solvefor y,(sqrt(y+1)-1)^2-2=x is y=x+2+2sqrt(x+2),y=x+2-2sqrt(x+2)