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Find the value of $p$ if the equation $(2p + 1)x^2 - (7p+2)x + 7p-3=0$ has real and equal roots.
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Sol. Given equation has real and equal roots if
$\{-(7p + 2)\}^2 - 4(2p + 1)(7p - 3) = 0$
$\Rightarrow 7p^2 - 24p - 16 = 0$
$\Rightarrow (7p + 4) (p - 4) = 0$
$\Rightarrow p = -\frac{4}{7}, p = 4$
$\{-(7p + 2)\}^2 - 4(2p + 1)(7p - 3) = 0$
$\Rightarrow 7p^2 - 24p - 16 = 0$
$\Rightarrow (7p + 4) (p - 4) = 0$
$\Rightarrow p = -\frac{4}{7}, p = 4$