# How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given 3/4x^2-1/3x-1=0?

May 9, 2018

$\Delta = \frac{1}{9} + 3 = \frac{28}{9} > 0 \implies \text{two distinct real roots}$

$x = \frac{2}{9} + \frac{4}{9} \sqrt{7}$
or
$x = \frac{2}{9} - \frac{4}{9} \sqrt{7}$

#### Explanation:

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

the discriminant is given by

$\Delta = {b}^{2} - 4 a c - - - \left(1\right)$

$\Delta > 0 \implies$two distinct real roots

$\Delta = 0 \implies$one root (ie two equal roots)

$\Delta < 0 \implies$ two distinct complex roots( conjugate pairs if $a , b , c \in \mathbb{R}$)

we have

$\frac{3}{4} {x}^{2} - \frac{1}{3} x - 1 = 0$

multiply the eqn by$12$

$a = \frac{3}{4} , b = - \frac{1}{3} , c = - 1$

$\left(1\right) \rightarrow \Delta = {\left(- \frac{1}{3}\right)}^{3} - 4 \left(\frac{3}{4}\right) \left(- 1\right)$

$\Delta = \frac{1}{9} + 3 = \frac{28}{9} > 0$

:. two distinct real roots

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$
$x = \frac{\frac{1}{3} \pm \sqrt{\frac{28}{9}}}{\frac{3}{2}}$
$x = \frac{\frac{1}{3} \pm \frac{2 \sqrt{7}}{3}}{\frac{3}{2}}$
$x = \frac{2}{9} \pm \frac{4}{9} \sqrt{7}$