# How do you use the discriminant to determine the numbers of solutions of the quadratic equation 2x^2-6x+5 = 0 and whether the solutions are real or complex?

Dec 23, 2016

$\Delta = - 4 < 0$, so this quadratic has a pair of non-Real Complex solutions.

#### Explanation:

$2 {x}^{2} - 6 x + 5 = 0$

is in the form:

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

with $a = 2$, $b = - 6$ and $c = 5$

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

$\Delta = {b}^{2} - 4 a c = {\left(- 6\right)}^{2} - 4 \left(2\right) \left(5\right) = 36 - 40 = - 4$

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

We can find the solutions by completing the square:

$0 = 2 \left(2 {x}^{2} - 6 x + 5\right)$

$\textcolor{w h i t e}{0} = 4 {x}^{2} - 12 x + 10$

$\textcolor{w h i t e}{0} = 4 {x}^{2} - 12 x + 9 + 1$

$\textcolor{w h i t e}{0} = {\left(2 x - 3\right)}^{2} - {i}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 x - 3\right) - i\right) \left(\left(2 x - 3\right) + i\right)$

$\textcolor{w h i t e}{0} = \left(2 x - 3 - i\right) \left(2 x - 3 + i\right)$

Hence solutions:

$x = \frac{3}{2} + \frac{1}{2} i \text{ }$ and $\text{ } x = \frac{3}{2} - \frac{1}{2} i$