How do you find the zeros, real and imaginary, of  y = 6x^2 +2x + 6  using the quadratic formula?

Dec 10, 2015

There are two imaginary roots at
$x = - \frac{1}{6} + \frac{\sqrt{35}}{6} i$ and $x = - \frac{1}{6} - \frac{\sqrt{35}}{6} i$

Explanation:

$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
has zeroes at
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Given:
$\textcolor{w h i t e}{\text{XXX}} y = 6 {x}^{2} + 2 x + 6$
$a = 6$
$b = 2$
$c = 6$

So the zeroes are at
color(white)("XXX")x=(-2+-sqrt(2^2-4(6)(6)))/(2(6)

$\textcolor{w h i t e}{\text{XXXx}} = \frac{- 2 \pm 2 \sqrt{1 - 36}}{2 \left(6\right)}$

$\textcolor{w h i t e}{\text{XXXx}} = - \frac{1 \pm \sqrt{- 35}}{6}$

$\textcolor{w h i t e}{\text{XXXx}} = - \frac{1}{6} \pm \frac{\sqrt{35}}{6} \cdot i$