# How do you find the zeros of g(x) = 4x^2-2x-3 using the quadratic formula?

Oct 26, 2016

$x = \frac{1}{4} \pm \frac{\sqrt{13}}{4}$

#### Explanation:

$g \left(x\right) = 4 {x}^{2} - 2 x - 3$

is in the standard form $a {x}^{2} + b x + c$, with $a = 4$, $b = - 2$ and $c = - 3$

We can find the zeros of $g \left(x\right)$ using the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(4\right) \left(- 3\right)}}{2 \cdot 4}$

$\textcolor{w h i t e}{x} = \frac{2 \pm \sqrt{4 + 48}}{8}$

$\textcolor{w h i t e}{x} = \frac{2 \pm \sqrt{52}}{8}$

$\textcolor{w h i t e}{x} = \frac{2 \pm \sqrt{{2}^{2} \cdot 13}}{8}$

$\textcolor{w h i t e}{x} = \frac{2 \pm 2 \sqrt{13}}{8}$

$\textcolor{w h i t e}{x} = \frac{1}{4} \pm \frac{\sqrt{13}}{4}$

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

Given any quadratic equation in the form $a {x}^{2} + b x + c = 0$,

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

is very useful, but do you know where it comes from, how it works or how to derive it yourself?

Here's one way...

Given $a {x}^{2} + b x + c = 0$, (with $a \ne 0$) we have:

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

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

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

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

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

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

Hence:

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