Question

Find the values of for which the quadratic equation ( + 1)

2 − 6( + 1) + 3( + 9) = 0 has equal roots. Hence, find the roots of the equation.​

Answer 1

Step-by-step explanation:

The given quadratic equation is

(p+1)x2−6(p+1)x+3(p+9)=0,p=−1

Compare given equation with the general form of quadratic equation, which ax2+bx+c=0

a=(p+1),b=−6(p+1) and c=3(p+9)

Discriminant:

D=b2−4ac

=(−6(p+1))2−4.(p+1).3(p+9)

=36(p+1)(p+1)−12(p+1)(p+6)

+12(p+1)(3p+3−p−9)

=12(p+1)(2p−6)

Since roots are real and equal (given)

Put D=0

12(p+1)(2p−6)=0

either (p+1)=0 or (2p−6)=0

p=−1 or p=3