# How do you find the roots of x^2-6x+12=0?

Mar 20, 2018

$x = 3 \pm \sqrt{3} \textcolor{w h i t e}{. .} i$

#### Explanation:

Given: $0 = {x}^{2} - 6 x + 12$

$\textcolor{b r o w n}{\text{The question specifically equates to 0. So we must find a solution that}}$color(brown)("works for "y=0.

Consider $y = a {x}^{2} + b x + c \textcolor{w h i t e}{\text{ddd")and color(white)("ddd}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Now lets look at the equation part : ${b}^{2} - 4 a c$. THis is called the determinant.

b^2-4ac color(white)("ddd")->color(white)("d") color(white)("ddd") (-6)^2-4(1)(12) = -12

As this is negative the graph does not have any x-intercepts where $x$ is in the set of what is called real numbers $x \notin \mathbb{R}$

There will be a solution of $x$ where it is in the set of 'Complex Numbers. $x \in \mathbb{C}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \textcolor{w h i t e}{\text{d")->color(white)("d}} x = \frac{+ 6 \pm \sqrt{- 12}}{2}$

$x = 3 \pm \sqrt{\frac{- 12}{4}}$

$x = 3 \pm \sqrt{3 \times \left(- 1\right)}$

$x = 3 \pm \sqrt{3} \times \sqrt{- 1}$

But $\sqrt{- 1} = i$

$x = 3 \pm \sqrt{3} \textcolor{w h i t e}{. .} i$ 