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Find the values of p for which the quadratic equation (p+1) x^{2}-6(p+1) x+3(p+9)=0. , p \neq-1 has equal roots. Hence find the roots of the equation.
CBSE | Class 10 | Exercise 10D | Maths | Quadratic Equations | RS Aggarwal
Find the values of p for which the quadratic equation (p+1) x^{2}-6(p+1) x+3(p+9)=0. , p \neq-1 has equal roots. Hence find the roots of the equation.

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).

\begin{array}{l} \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) \end{array}

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

\begin{array}{l} \therefore 12(p+1)(2 p-6)=0 \\ \Rightarrow p+1=0 \text { or } 2 p-6=0 \\ \Rightarrow p=-1 \text { or } p=3 \end{array}

But, p \neq-1 \quad (Given)

Thus, the value of p is 3

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

\begin{array}{l} 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 \end{array}

Hence, 3 is the repeated root of this equation.

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