How do you find the roots for 20x^2 + 13x + 2?

Feb 19, 2016

Either factor or use the quadratic formula to obtain
$x \in \left\{- \frac{1}{4} , - \frac{2}{5}\right\}$

Explanation:

Using the quadratic formula (this can sometimes be faster if the factors are not obvious)
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
in this case
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- 13 \pm \sqrt{169 - 160}}{40} = \frac{- 13 \pm 3}{40}$

$\textcolor{w h i t e}{\text{XXX}} x = - \frac{10}{40} = - \frac{1}{4}$ or $x = - \frac{16}{40} = - \frac{2}{5}$

Factoring
$20 {x}^{2} + 13 x + 2 = \left(4 x + 1\right) \left(5 x + 2\right)$
The roots are the values of $x$ for which the expression $= 0$
So
$\textcolor{w h i t e}{\text{XXX}} \left(4 x + 1\right) = 0 \rightarrow x = - \frac{1}{4}$
or
$\textcolor{w h i t e}{\text{XXX}} \left(5 x + 2\right) = 0 \rightarrow x = - \frac{2}{5}$