The values of p for which the equation has real roots is/are ____.(2p + 1)x2-7p+2x+7p-3=0

# The values of p for which the equation has real roots is/are ____.(2p + 1)${x}^{2}-\left(7p+2\right)x+\left(7p-3\right)=0$

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### Solution:

(2p + 1)${x}^{2}-\left(7p+2\right)x+\left(7p-3\right)=0$
On comparing the equation with $a{x}^{2}+\mathit{bx}+c=0$, we get
a = 2p + 1, b = - (7p + 2) and c = 7p - 3   Since, the roots are equal and real
${b}^{2}-4\mathit{ac}=0$ ${\left(7p+2\right)}^{2}-4\left(2p+1\right)\left(7p-3\right)$ = 0
49${p}^{2}+4+28p-4\left(14{p}^{2}-6p+7p-3\right)=0$
49${p}^{2}+4+28p$ - 56– 4p + 12 = 0
-7${p}^{2}+24p+16=0$
$7{p}^{2}-24p-16=0$ ${7p}^{2}-28p+4p-16=0$ 7p(p – 4) + 4(p - 4) = 0
(7p + 4) = 0 and (p – 4) = 0
p = - $\frac{4}{7}$ and 4

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