Find the values of p for which the quadratic equation
Question:

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

 

Solution:

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 $c=7 p-3$.

$\therefore D=b^{2}-4 a c$

$=[-(7 p+2)]^{2}-4 \times(2 p+1) \times(7 p-3)$

$=\left(49 p^{2}+28 p+4\right)-4\left(14 p^{2}+p-3\right)$

$=49 p^{2}+28 p+4-56 p^{2}-4 p+12$

$=-7 p^{2}+24 p+16$

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

$\therefore-7 p^{2}+24 p+16=0$

$\Rightarrow 7 p^{2}-24 p-16=0$

$\Rightarrow 7 p^{2}-28 p+4 p-16=0$

$\Rightarrow 7 p(p-4)+4(p-4)=0$

$\Rightarrow(p-4)(7 p+4)=0$

$\Rightarrow p-4=0$ or $7 p+4=0$

$\Rightarrow p=4$ or $p=-\frac{4}{7}$

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

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