# How do you find the roots, real and imaginary, of y= 12x^2 - 12x + 8+ 4(-x-1)^2  using the quadratic formula?

Sep 16, 2016

Two imaginary roots.
$x = \frac{1 \pm i \sqrt{47}}{8}$

#### Explanation:

Develop and bring the equation to standard form:
$4 {\left(- x - 1\right)}^{2} = 4 {\left[- \left(x + 1\right)\right]}^{2} = 4 {\left(x + 1\right)}^{2} = 4 \left({x}^{2} + 2 x + 1\right) = 4 {x}^{2} + 8 x + 4$.
$y = 12 {x}^{2} - 12 x + 8 + \left(4 {x}^{2} + 8 x + 4\right) = 0$
$y = 16 {x}^{2} - 4 x + 12 = 4 \left(4 {x}^{2} - x + 3\right) = 0$
$y = 4 {x}^{2} - x + 3 = 0$
$D = 1 - 48 = - 47 < 0$ --> $d = \pm i \sqrt{47}$
Since D < 0, there are 2 imaginary roots.
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{1}{8} \pm \frac{i \sqrt{47}}{8} = \frac{1 \pm i \sqrt{47}}{8}$

Sep 16, 2016

$x = \frac{1}{8} \pm \frac{\sqrt{47}}{8} \textcolor{w h i t e}{.} i$

#### Explanation:

Given:$\text{ } y = 12 {x}^{2} - 12 x + 8 + 4 {\left(- x - 1\right)}^{2}$

Squaring the bracket

$y = 12 {x}^{2} - 12 x + 8 + 4 \left({x}^{2} + 2 x + 1\right)$

giving:

$y = 12 {x}^{2} - 12 x + 8 + 4 {x}^{2} + 8 x + 4$

$y = 16 {x}^{2} - 4 x + 12 \textcolor{red}{\leftarrow \text{ slightly different approach hear.}}$

Set $y = 0$

$\implies 0 = 16 {x}^{2} - 4 x + 12$

divide both sides by 4

Note that $\frac{0}{4} = 0$

$\implies 0 = 4 {x}^{2} - x + 3$

Compare to $y = a {x}^{2} + b x + c \text{ "->" } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Where$\text{ "a=4"; "b=-1"; } c = 3$

$\implies x = \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(4\right) \left(3\right)}}{2 \left(4\right)}$

$x = \frac{1 \pm \sqrt{1 - 48}}{8}$

$x = \frac{1}{8} \pm \frac{\sqrt{- 47}}{8}$

But 47 is a prime number so we have

$x = \frac{1}{8} \pm \frac{\sqrt{47}}{8} \textcolor{w h i t e}{.} i$