Question

If 3 is a root of the quadratic equation $x^2-x+k=0$, find the value of p so that the roots of the equation $x^2+k\left(2x+k+2\right)+p=0$ are equal.
[CBSE 2015]

Answer 1

It is given that 3 is a root of the quadratic equation $x^2-x+k=0$.

$$\begin{gathered} \therefore {\left(3\right)}^2-3+k=0 \\ \Rightarrow k+6=0 \\ \Rightarrow k=-6 \end{gathered}$$

The roots of the equation $x^2+2kx+\left(k^2+2k+p\right)=0$ are equal.

$$\begin{gathered} \therefore D=0 \\ \Rightarrow {\left(2k\right)}^2-4\times 1\times \left(k^2+2k+p\right)=0 \\ \Rightarrow 4k^2-4k^2-8k-4p=0 \\ \Rightarrow -8k-4p=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow p=\frac{8k}{-4}=-2k \\ \Rightarrow p=-2\times \left(-6\right)=12 \end{gathered}$$

Hence, the value of p is 12.