Find the value of $\mathrm{p}$ for which the quadratic equation $(2 p+1) x^{2}-(7 p+2) x+(7 p-3)=0$. has real and equal roots.

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Find the value of $\mathrm{p}$ for which the quadratic equation $(2 p+1) x^{2}-(7 p+2) x+(7 p-3)=0$ . has real and equal roots.

Find the value of $\mathrm{p}$ for which the quadratic equation $(2 p+1) x^{2}-(7 p+2) x+(7 p-3)=0$ . has real and equal roots.

The given equation is $(2 p+1) x^{2}-(7 p+2) x+(7 p-3)=0$ .

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

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

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

$\begin{array}{l} \therefore-7 p^{2}+24 p+16=0 \\ \Rightarrow 7 p^{2}-24 p-16=0 \\ \Rightarrow 7 p^{2}-28 p+4 p-16=0 \\ \Rightarrow 7 p(p-4)+4(p-4)=0 \\ \Rightarrow(p-4)(7 p+4)=0 \\ \Rightarrow p-4=0 \text { or } 7 p+4=0 \end{array}$