# How do you solve the equation 1/7x^2-3=4?

Apr 3, 2017

$x = \pm 7$

#### Explanation:

Start by adding 3 to both sides of the equation.

$\frac{1}{7} {x}^{2} \cancel{- 3} \cancel{+ 3} = 4 + 3$

$\Rightarrow \frac{1}{7} {x}^{2} = 7$

multiply both sides by 7 to eliminate the fraction.

${\cancel{7}}^{1} \times \frac{1}{\cancel{7}} ^ 1 {x}^{2} = 7 \times 7$

$\Rightarrow {x}^{2} = 49$

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

$\sqrt{{x}^{2}} = \pm \sqrt{49}$

$\Rightarrow x = \pm 7 \leftarrow \textcolor{red}{\text{ are the solutions}}$