What are the solutions to the equation solvefor x,sqrt(y)=(x+4)^2 ?

The solutions to the equation solvefor x,sqrt(y)=(x+4)^2 are x=\sqrt[4]{y}-4,x=-\sqrt[4]{y}-4

Find the zeros of solvefor x,sqrt(y)=(x+4)^2

The zeros of solvefor x,sqrt(y)=(x+4)^2 are x=\sqrt[4]{y}-4,x=-\sqrt[4]{y}-4

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solve for

x, y = (x + 4)^{2}

Solution steps

y = (x + 4)^{2}

Switch sides

(x + 4)^{2} = y

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

The solutions to the quadratic equation are:

9 w^{2} + 30 w = - 25

- 4.9 x^{2} + 4.3301200000000 \dots x - 0.8 = 0

x^{2} + 2 x - 21 = 0

4 k^{2} - 9 = 0

6 x^{2} - x - 2 = 0

What are the solutions to the equation solvefor x,sqrt(y)=(x+4)^2 ?
The solutions to the equation solvefor x,sqrt(y)=(x+4)^2 are x=\sqrt[4]{y}-4,x=-\sqrt[4]{y}-4
Find the zeros of solvefor x,sqrt(y)=(x+4)^2
The zeros of solvefor x,sqrt(y)=(x+4)^2 are x=\sqrt[4]{y}-4,x=-\sqrt[4]{y}-4