# How do you find the value of the discriminant and determine the nature of the roots 6p^2-2p-3?

Jul 18, 2017

If we call the coefficients $A , B , C$ respectively, the discriminant is:
$D = {B}^{2} - 4 A C$

#### Explanation:

$D = {\left(- 2\right)}^{2} - 4 \cdot 6 \cdot \left(- 3\right) = 4 + 72 = 76$

This is positive ($D > 0$), so there are two distinct real roots:

${x}_{1 , 2} = \frac{- B \pm \sqrt{D}}{2 A} = \frac{2 \pm \sqrt{76}}{2 \cdot 6} = \frac{\cancel{2} 1 \pm \cancel{2} \sqrt{19}}{\cancel{2} \cdot 6}$

${x}_{1 , 2} = \frac{1}{6} \pm \frac{1}{6} \sqrt{19}$

Jul 18, 2017

Discriminant is positive , so there are two real roots.

#### Explanation:

$6 {p}^{2} - 2 p - 3$ Comparing with standard quadratic equation

ax^2+bx+c ; a= 6 , b= -2 ,c =-3

Discriminant $D = {b}^{2} - 4 \cdot a \cdot c = {\left(- 2\right)}^{2} - 4 \cdot 6 \cdot \left(- 3\right) = 4 + 72 = 76$

$D > 0$ . If $D$ is positive, we get two real roots, if it is zero we

get just one root, and if it is negative we get complex roots.

Here $D$ is positive , so there are two real roots. [Ans]