# How do you find the discriminant and how many solutions does 2j^2 - 3j = -1 have?

May 10, 2015

discriminant $D = {b}^{2} - 4 a c$

we have: $2 {j}^{2} - 3 j + 1 = 0$

here:
$a = 2$ , $b = - 3$, $c = 1$
(coefficients of ${j}^{2}$ , $j$ and the constant term respectively)

finding $D$:
$D = {b}^{2} - 4 a c = \left(- {3}^{2}\right) - \left(4 \times 2 \times 1\right)$
$D = 9 - 8 = 1$

formula for roots :
$j = \frac{- b \pm \sqrt{D}}{2 a}$
$j = \frac{3 \pm \sqrt{1}}{2 \times 2}$
 j = (3 + 1) / 4 = 4/4 = 1 and (3 -1) /4 = 2/4 = 1/2
$j$ has two solutions:
$j = 1$ and $j = \frac{1}{2}$

the roots are real and unequal.