# How do you solve 9x^2=25 using the quadratic formula?

Jan 16, 2017

See entire solution process below:

#### Explanation:

First, we need to transform this into quadratic form by subtracting $\textcolor{red}{25}$ from each side of the equation to keep the equation balanced while equating to $0$:

$9 {x}^{2} - \textcolor{red}{25} = 25 - \textcolor{red}{25}$

$9 {x}^{2} - 25 = 0$

This is a special form of the quadratic equation which has the solution:

$\textcolor{red}{a} {x}^{2} - \textcolor{b l u e}{b} = \left(\sqrt{\textcolor{red}{a}} x + \sqrt{\textcolor{b l u e}{b}}\right) \left(\sqrt{\textcolor{red}{a}} x - \sqrt{\textcolor{b l u e}{b}}\right) = 0$

$\textcolor{red}{9} {x}^{2} - \textcolor{b l u e}{25} = \left(\sqrt{\textcolor{red}{9}} x + \sqrt{\textcolor{b l u e}{25}}\right) \left(\sqrt{\textcolor{red}{9}} x - \sqrt{\textcolor{b l u e}{25}}\right) = 0$

$\left(3 x + 5\right) \left(3 x - 5\right) = 0$

or, because the $\sqrt{9} = \pm 3$

$\left(- 3 x + 5\right) \left(- 3 x - 5\right) = 0$

Now, we can solve each term for $0$:

Solution 1)

$3 x + 5 = 0$

$3 x + 5 - 5 = 0 - 5$

$3 x + 0 = - 5$

$3 x = - 5$

$\frac{3 x}{3} = - \frac{5}{3}$

$x = - \frac{5}{3}$

Solution 2)

$3 x - 5 = 0$

$3 x - 5 + 5 = 0 + 5$

$3 x + 0 = 5$

$3 x = 5$

$\frac{3 x}{3} = \frac{5}{3}$

$x = \frac{5}{3}$

Solution 3)

$- 3 x + 5 = 0$

$- 3 x + 5 - 5 = 0 - 5$

$- 3 x + 0 = - 5$

$- 3 x = - 5$

$\frac{- 3 x}{- 3} = \frac{- 5}{-} 3$

$x = \frac{5}{3}$

Solution 4)

$- 3 x - 5 = 0$

$- 3 x - 5 + 5 = 0 + 5$

$- 3 x + 0 = 5$

$- 3 x = 5$

$\frac{- 3 x}{-} 3 = \frac{5}{-} 3$

$x = - \frac{5}{3}$

$x = \frac{5}{3}$ or $x = - \frac{5}{3}$