Question

Find the values of p for which the quadratic equation $\left(2p+1\right)x^2-\left(7p+2\right)x+\left(7p-3\right)=0$ has real and equal roots. [CBSE 2014]

Answer 1

The given equation is $\left(2p+1\right)x^2-\left(7p+2\right)x+\left(7p-3\right)=0$.

This is of the form $a{x}^2+bx+c=0$, where a = 2p +1, b = −(7p + 2) and c = 7p − 3.

$$\begin{gathered} \therefore D=b^2-4ac \\ ={\left[-\left(7p+2\right)\right]}^2-4\times \left(2p+1\right)\times \left(7p-3\right) \\ =\left(49p^2+28p+4\right)-4\left(14p^2+p-3\right) \\ =49p^2+28p+4-56p^2-4p+12 \end{gathered}$$
$=-7p^2+24p+16$

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

$$\begin{gathered} \therefore -7p^2+24p+16=0 \\ \Rightarrow 7p^2-24p-16=0 \\ \Rightarrow 7p^2-28p+4p-16=0 \\ \Rightarrow 7p\left(p-4\right)+4\left(p-4\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(p-4\right)\left(7p+4\right)=0 \\ \Rightarrow p-4=0\mathrm{or}7p+4=0 \\ \Rightarrow p=4\mathrm{or}p=-\frac{4}{7} \end{gathered}$$

Hence, 4 and $-\frac{4}{7}$ are the required values of p.