We know that while finding the root of a quadratic equation ax2+bx+c=0 by quadratic formula x=−b±√b2−4ac2a,
if b2−4ac>0, then the roots are real and distinct
if b2−4ac=0, then the roots are real and equal and
if b2−4ac<0, then the roots are imaginary.
Here, the given quadratic equation (p+1)n2+2(p+3)n+(p+8)=0 is in the form ax2+bx+c=0 where a=(p+1),b=2(p+3)=(2p+6) and c=(p+8).
It is given that the roots are equal, therefore b2−4ac=0 that is:
b2−4ac=0⇒(2p+6)2−(4×(p+1)×(p+8))=0⇒(2p)2+62+(2×2p×6)−4[p(p+8)+1(p+8)]=0⇒(4p2+36+24p)−4(p2+8p+p+8)=0⇒(4p2+36+24p)−4(p2+9p+8)=0⇒(4p2+36+24p)−4p2−36p−32=0⇒4p2+36+24p−4p2−36p−32=0⇒−12p+4=0⇒−12p=−4⇒p=412⇒p=13
Hence, p=13