Question

If −4 is a root of the quadratic equation $x^2+2x+4p=0$, find the value of k for which the quadratic equation $x^2+px\left(1+3k\right)+7\left(3+2k\right)=0$ has equal roots. [CBSE 2015]

Answer 1

It is given that −4 is a root of the quadratic equation $x^2+2x+4p=0$.

$$\begin{gathered} \therefore {\left(-4\right)}^2+2\times \left(-4\right)+4p=0 \\ \Rightarrow 16-8+4p=0 \\ \Rightarrow 4p+8=0 \\ \Rightarrow p=-2 \end{gathered}$$

The equation $x^2+px\left(1+3k\right)+7\left(3+2k\right)=0$ has equal roots.

$$\begin{gathered} \therefore D=0 \\ \Rightarrow {\left[p\left(1+3k\right)\right]}^2-4\times 1\times 7\left(3+2k\right)=0 \\ \Rightarrow {\left[-2\left(1+3k\right)\right]}^2-28\left(3+2k\right)=0 \\ \Rightarrow 4\left(1+6k+9k^2\right)-28\left(3+2k\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 4\left(1+6k+9k^2-21-14k\right)=0 \\ \Rightarrow 9k^2-8k-20=0 \\ \Rightarrow 9k^2-18k+10k-20=0 \\ \Rightarrow 9k\left(k-2\right)+10\left(k-2\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(k-2\right)\left(9k+10\right)=0 \\ \Rightarrow k-2=0\mathrm{or}9k+10=0 \\ \Rightarrow k=2\mathrm{or}k=-\frac{10}{9} \end{gathered}$$

Hence, the required value of k is 2 or $-\frac{10}{9}$.