# How do you solve 25v^2=1?

Sep 4, 2016

$v = \pm \frac{1}{5}$

#### Explanation:

Divide both sides by 25 giving:

${v}^{2} = \frac{1}{25}$

Thus:

$\sqrt{{v}^{2}} = \sqrt{\frac{1}{25}}$

$v = \frac{\sqrt{1}}{\sqrt{25}} = \frac{1}{\sqrt{25}}$

But as this is square root we should write $= \pm \frac{1}{\sqrt{25}}$

$v = \pm \frac{1}{5}$

Sep 4, 2016

$v = \pm \frac{1}{5}$

#### Explanation:

Begin by isolating ${v}^{2}$. To do this we divide both sides of the equation by 25.

$\Rightarrow \frac{{\cancel{25}}^{1} {v}^{2}}{\cancel{25}} ^ 1 = \frac{1}{25} \Rightarrow {v}^{2} = \frac{1}{25}$

now take the $\textcolor{b l u e}{\text{square root of both sides}}$

$\Rightarrow \sqrt{{v}^{2}} = \pm \sqrt{\frac{1}{25}}$

$\Rightarrow v = \pm \frac{1}{5}$

Sep 4, 2016

$v = \pm \frac{1}{5}$

#### Explanation:

Apart from the method shown by other contributors, we can also follow the usual method for a quadratic equation.

$\rightarrow \text{Make it equal to 0 " rarr " find the factors.}$

$25 {v}^{2} = 1$

$25 {v}^{2} - 1 = 0 \leftarrow \text{ difference of squares}$

$\left(5 v + 1\right) \left(5 v - 1\right) = 0$

Either factor could be 0

$5 v + 1 = 0 \text{ } \rightarrow v = - \frac{1}{5}$

$5 v - 1 = 0 \text{ } \rightarrow v = \frac{1}{5}$