# How to use the discriminant to find out how many real number roots an equation has for 4/3x^2 - 2x + 3/4 = 0?

May 16, 2015

$\left(\frac{4}{3}\right) {x}^{2} - 2 x + \frac{3}{4}$ is of the form $a {x}^{2} + b x + c$

with $a = \frac{4}{3}$, $b = - 2$ and $c = \frac{3}{4}$.

The discriminant $\Delta$ is given by the formula ${b}^{2} - 4 a c$

In our case:

$\Delta = {b}^{2} - 4 a c$

$= {\left(- 2\right)}^{2} - 4 \cdot \left(\frac{4}{3}\right) \cdot \left(\frac{3}{4}\right)$

$= 4 - 4 = 0$

If $\Delta > 0$ then our equation would have 2 real roots.

If $\Delta < 0$ then our equation would have 2 complex non-real roots.

In our case $\Delta = 0$ implies that our equation has 1 repeated real root.

That root is

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- \left(- 2\right) \pm 0}{2 \cdot \left(\frac{4}{3}\right)} = \frac{2}{2 \cdot \left(\frac{4}{3}\right)} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$