What is the answer to solvefor y,x^2= 1/(sqrt(y+4)) ?

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

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solvefor y,x^2= 1/(sqrt(y+4))

solve for

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

y = \frac{1}{x ^{4}} - 4

Solution steps

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

Multiply both sides by

y + 4

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

Simplify

x^{2} y + 4 = 1

Divide both sides by

x^{2}

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

Square both sides:

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

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

Solve

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

y = \frac{1}{x ^{4}} - 4

Verify Solutions:

y = \frac{1}{x ^{4}} - 4

True

The solution is

y = \frac{1}{x ^{4}} - 4

h^{2} + 400 h - 125000 = 0

u^{2} = - 16

3 x^{2} + 5 x - 322 = 0

(7 x + 6)^{2} = 16

(- 0.2) x^{2} + 80 x - 3.95 = 0

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