# How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given x^2-x+6=0?

Jan 5, 2017

$x = \frac{1}{2} \pm \sqrt{23} i$

#### Explanation:

The discriminant $\Delta$ of a quadratic equation with rational coefficients gives us the following information:

• If $\Delta > 0$ then there are two, distinct, real roots. If $\Delta$ is a perfect square, then they are both rational.

• If $\Delta = 0$ then there is one, repeated, rational, real root.

• If $\Delta < 0$ then there are two, distinct, non-Real, Complex roots, which form a complex conjugate pair.

${x}^{2} - x + 6 = 0$

is in the form:

$a {x}^{2} + b x + c = 0$

with $a = 1$, $b = - 1$ and $c = 6$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{- 1}\right)}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{6}\right) = 1 - 24 = - 23$

Since $\Delta < 0$ this quadratic has no Real roots. It has a complex conjugate pair of non-Real, Complex roots.

We can find the roots using the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{1 \pm \sqrt{- 23}}{2 \cdot 1}$

$\textcolor{w h i t e}{x} = \frac{1 \pm \sqrt{23} i}{2}$

$\textcolor{w h i t e}{x} = \frac{1}{2} \pm \sqrt{23} i$