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

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Answer 1 · 2017-07-18T12:10:22

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

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

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

$x_(1,2)=(-B+-sqrtD)/(2A)=(2+-sqrt76)/(2*6)=(cancel2 1+-cancel2sqrt19)/(cancel2*6)$

$x_(1,2)=1/6+-1/6sqrt19$

Answer 2 · 2017-07-18T12:11:54

Discriminant is positive , so there are two real roots.

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

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

Discriminant $D = b^2- 4*a*c = (-2)^2 -4 * 6 * (-3) = 4+72=76$

$D > 0$ . 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]

How do you find the value of the discriminant and determine the nature of the roots 6p^2-2p-36p2−2p−36p^2-2p-3?