# How do you use the quadratic formula to solve the equation, x^2-x= -1?

Oct 18, 2016

NO ROOTS in $x \notin \mathbb{R}$
ROOTS $x \in \mathbb{C}$
$x = \frac{1 + i \sqrt{3}}{2}$
OR
$x = \frac{1 - i \sqrt{3}}{2}$

#### Explanation:

${x}^{2} - x = - 1$
$\Rightarrow {x}^{2} - x + 1 = 0$
We have to factorize
$\textcolor{b r o w n}{{x}^{2} - x + 1}$
Since we can not use polynomial identities so we will calculate $\textcolor{b l u e}{\delta}$

$\textcolor{b l u e}{\delta = {b}^{2} - 4 a c}$
$\delta = {\left(- 1\right)}^{2} - 4 \left(1\right) \left(1\right) = - 3 < 0$

NO ROOTS IN $\textcolor{red}{x \notin \mathbb{R}}$ because $\textcolor{red}{\delta < 0}$

But roots exist in $\mathbb{C}$
$\textcolor{b l u e}{\delta = 3 {i}^{2}}$

Roots are
${x}_{1} = \frac{- b + \sqrt{\delta}}{2 a} = \frac{1 + \sqrt{3 {i}^{2}}}{2} = \frac{1 + i \sqrt{3}}{2}$
${x}_{2} = \frac{- b - \sqrt{\delta}}{2 a} = \frac{1 - \sqrt{3 {i}^{2}}}{2} = \frac{1 - i \sqrt{3}}{2}$

The equation is:
${x}^{2} - x + 1 = 0$
$\Rightarrow \left(x - \frac{1 + i \sqrt{3}}{2}\right) \left(x - \frac{1 - i \sqrt{3}}{2}\right) = 0$
$\left(x - \frac{1 + i \sqrt{3}}{2}\right) = 0 \Rightarrow \textcolor{b r o w n}{x = \frac{1 + i \sqrt{3}}{2}}$
OR
$\left(x - \frac{1 - i \sqrt{3}}{2}\right) = 0 \Rightarrow \textcolor{b r o w n}{x = \frac{1 - i \sqrt{3}}{2}}$

So the roots exist only in $\textcolor{red}{x \in \mathbb{C}}$