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

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Answer 1 · 2017-07-18T12:10:22

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

#D=(-2)^2-4*6*(-3)=4+72=76#

This is positive (#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]