Question

Quadratic equation (P +1)x² – 2(P-1)x+1=0 have real and equal roots, then find value of P.​

Answer 1

Answer:

The given quadratic equation is

(p+1)x

2

−6(p+1)x+3(p+9)=0,p



=−1

Compare given equation with the general form of quadratic equation, which ax

2

+bx+c=0

a=(p+1),b=−6(p+1) and c=3(p+9)

Discriminant:

D=b

2

−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