Question

Find the value of p for which the quadratic equation $\left(p+1\right)x^2-6(p+1)x+3(p+9)=0,p\ne -1$ has equal roots. Hence, find the roots of the equation.

Disclaimer: There is a misprinting in the given question. In the question 'q' is printed instead of 9.

Answer 1

The given quadratic equation $\left(p+1\right)x^2-6(p+1)x+3(p+9)=0$, has equal roots.

Here, $a=p+1,b=-6p-6\mathrm{and}c=3p+27$.

As we know that $D=b^2-4ac$

Putting the values of $a=p+1,b=-6p-6\mathrm{and}c=3p+27$.

$$\begin{gathered} D={\left[-6(p+1)\right]}^2-4\left(p+1\right)\left[3\left(p+9\right)\right] \\ =36(p^2+2p+1)-12(p^2+10p+9) \\ =36p^2-12p^2+72p-120p+36-108 \\ =24p^2-48p-72 \end{gathered}$$

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

Thus, $24p^2-48p-72=0$
$$\begin{gathered} \Rightarrow p^2-2p-3=0 \\ \Rightarrow p^2-3p+p-3=0 \\ \Rightarrow p(p-3)+1(p-3)=0 \\ \Rightarrow (p+1)(p-3)=0 \\ \Rightarrow p+1=0\mathrm{or}p-3=0 \\ \Rightarrow p=-1\mathrm{or}p=3 \end{gathered}$$

Therefore, the value of p is −1, 3.

It is given that p ≠ −1, thus p = 3 only.

Now the equation becomes

$$\begin{gathered} 4x^2-24x+36=0 \\ \Rightarrow x^2-6x+9=0 \\ \Rightarrow x^2-3x-3x+9=0 \\ \Rightarrow x(x-3)-3(x-3)=0 \\ \Rightarrow (x-3{)}^2=0 \\ \Rightarrow x=3,3 \end{gathered}$$

​Hence, the root of the equation is 3.