# How do you use the quadratic formula to solve 9x^2-6x-35=0?

May 7, 2017

$x = \frac{7}{3} \text{ }$ or $\text{ } x = - \frac{5}{3}$

#### Explanation:

Given:

$9 {x}^{2} - 6 x - 35 = 0$

Note that this is in standard form:

$a {x}^{2} + b x + c = 0$

with $a = 9$, $b = - 6$ and $c = - 35$

The roots are given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- \left(\textcolor{b l u e}{- 6}\right) \pm \sqrt{{\left(\textcolor{b l u e}{- 6}\right)}^{2} - 4 \left(\textcolor{b l u e}{9}\right) \left(\textcolor{b l u e}{- 35}\right)}}{2 \left(\textcolor{b l u e}{9}\right)}$

$\textcolor{w h i t e}{x} = \frac{6 \pm \sqrt{36 + 1260}}{18}$

$\textcolor{w h i t e}{x} = \frac{6 \pm \sqrt{1296}}{18}$

$\textcolor{w h i t e}{x} = \frac{6 \pm 36}{18}$

$\textcolor{w h i t e}{x} = \frac{1 \pm 6}{3}$

That is:

$x = \frac{7}{3} \text{ }$ or $\text{ } x = - \frac{5}{3}$

$\textcolor{w h i t e}{}$
Notes

The quadratic formula is very useful and worth memorizing, but I would strongly recommend that you learn how to derive it from scratch - if you do not know already.

Here's one way, making use of the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

Given:

$a {x}^{2} + b x + c = 0$

We can write:

$0 = a {x}^{2} + b x + c$

$\textcolor{w h i t e}{0} = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

$\textcolor{w h i t e}{0} = a \left({\left(x + \frac{b}{2 a}\right)}^{2} - \frac{{b}^{2} - 4 a c}{2 a} ^ 2\right)$

$\textcolor{w h i t e}{0} = a \left({\left(x + \frac{b}{2 a}\right)}^{2} - {\left(\frac{\sqrt{{b}^{2} - 4 a c}}{2 a}\right)}^{2}\right)$

$\textcolor{w h i t e}{0} = a \left(\left(x + \frac{b}{2 a}\right) - \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(\left(x + \frac{b}{2 a}\right) + \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}\right)$

$\textcolor{w h i t e}{0} = a \left(x - \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(x - \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}\right)$

Hence:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$