Find the values of p for which the quadratic equation

Solution:

The given equation is $(p+1) x^{2}-6(p+1) x+3(p+9)=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=p+1, b=-6(p+1)$ and $c=3(p+9)$.

$\therefore D=b^{2}-4 a c$

$=[-6(p+1)]^{2}-4 \times(p+1) \times 3(p+9)$

$=12(p+1)[3(p+1)-(p+9)]$

$=12(p+1)(2 p-6)$

The given equation will have real and equal roots if D = 0.

$\therefore 12(p+1)(2 p-6)=0$

$\Rightarrow p+1=0$ or $2 p-6=0$

$\Rightarrow p=-1$ or $p=3$

But, $p \neq-1$      (Given)

Thus, the value of p is 3.

Putting $p=3$, the given equation becomes $4 x^{2}-24 x+36=0$.

$4 x^{2}-24 x+36=0$

$\Rightarrow 4\left(x^{2}-6 x+9\right)=0$

$\Rightarrow(x-3)^{2}=0$

$\Rightarrow x-3=0$

$\Rightarrow x=3$

Hence, 3 is the repeated root of this equation.