What are the solutions to the equation solvefor y,y^2=x^{x+2} ?

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

Find the zeros of solvefor y,y^2=x^{x+2}

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

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solve for

y, y^{2} = x^{x + 2}

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

Solution steps

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

For

x^{2} = f (a)

the solutions are

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

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

4 x + 4 \leq 9 x + 84 x + 4 \leq 9 x + 8

\frac{4 ( x + 5 )}{5} = 12

- \frac{12}{3}

\frac{a}{7} - 3 = - 2

2 x + 2 \leq 10

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