What is the answer to solvefor y,x=sqrt(2y-\sqrt{y)} ?

The answer to solvefor y,x=sqrt(2y-\sqrt{y)} is y=(4x^2+1+sqrt(8x^2+1))/8 ,y=(4x^2+1-sqrt(8x^2+1))/8

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solvefor y,x=sqrt(2y-\sqrt{y)}

solve for

y, x = 2 y - y

y = \frac{4 x ^{2} + 1 + 8 x ^{2} + 1}{8}, y = \frac{4 x ^{2} + 1 - 8 x ^{2} + 1}{8}

Solution steps

x = 2 y - y

Remove square roots

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

Solve

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

y = \frac{4 x ^{2} + 1 + 8 x ^{2} + 1}{8}, y = \frac{4 x ^{2} + 1 - 8 x ^{2} + 1}{8}

Verify Solutions:

y = \frac{4 x ^{2} + 1 + 8 x ^{2} + 1}{8}

True

, y = \frac{4 x ^{2} + 1 - 8 x ^{2} + 1}{8}

True

The solutions are

y = \frac{4 x ^{2} + 1 + 8 x ^{2} + 1}{8}, y = \frac{4 x ^{2} + 1 - 8 x ^{2} + 1}{8}

6 + x - x^{2} = 9

x^{- \frac{1}{2}} = x

x - 5 = x - 4

(2 x + 5)^{\frac{2}{3}} = 16

x (3 x - 7) - 32 - x = - 1

What is the answer to solvefor y,x=sqrt(2y-\sqrt{y)} ?
The answer to solvefor y,x=sqrt(2y-\sqrt{y)} is y=(4x^2+1+sqrt(8x^2+1))/8 ,y=(4x^2+1-sqrt(8x^2+1))/8