What are the solutions to the equation sqrt(128)=x^2 ?

The solutions to the equation sqrt(128)=x^2 are x=2\sqrt[4]{2^3},x=-2\sqrt[4]{2^3}

Find the zeros of sqrt(128)=x^2

The zeros of sqrt(128)=x^2 are x=2\sqrt[4]{2^3},x=-2\sqrt[4]{2^3}

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128 = x^{2}

Decimal Notation

x = 3.36358 \dots, x = - 3.36358 \dots

Solution steps

128 = x^{2}

Switch sides

x^{2} = 128

Simplify

128 : 82

For

x^{2} = f (a)

the solutions are

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

x = 82, x = - 82

3 x = - 27

2 x = - 8

\frac{8}{6} - \frac{1}{8}

x^{2} - 10 x + 21 \leq 0

(\frac{2}{3})^{3} + \frac{2}{9}

What are the solutions to the equation sqrt(128)=x^2 ?
The solutions to the equation sqrt(128)=x^2 are x=2\sqrt[4]{2^3},x=-2\sqrt[4]{2^3}
Find the zeros of sqrt(128)=x^2
The zeros of sqrt(128)=x^2 are x=2\sqrt[4]{2^3},x=-2\sqrt[4]{2^3}