Evaluate v^2=25/81

$v^{2} = \frac{25}{81}$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$v = \pm \sqrt{\frac{25}{81}}$

Simplify $\pm \sqrt{\frac{25}{81}}$ .

Rewrite $\sqrt{\frac{25}{81}}$ as $\frac{\sqrt{25}}{\sqrt{81}}$ .

$v = \pm \frac{\sqrt{25}}{\sqrt{81}}$

Simplify the numerator.

Rewrite $25$ as $5^{2}$ .

$v = \pm \frac{\sqrt{5^{2}}}{\sqrt{81}}$

Pull terms out from under the radical, assuming positive real numbers.

$v = \pm \frac{5}{\sqrt{81}}$

Simplify the denominator.

Rewrite $81$ as $9^{2}$ .

$v = \pm \frac{5}{\sqrt{9^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$v = \pm \frac{5}{9}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$v = \frac{5}{9}$

Next, use the negative value of the $\pm$ to find the second solution.

$v = - \frac{5}{9}$

The complete solution is the result of both the positive and negative portions of the solution.

$v = \frac{5}{9}, - \frac{5}{9}$

The result can be shown in multiple forms.

Exact Form:

$v = \frac{5}{9}, - \frac{5}{9}$

Decimal Form:

$v = 0.\overline{5}, - 0.\overline{5}$