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Find the value of $p$ for which the quadratic equation $(2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0$ has equal roots. Also, find these roots.
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For equal roots, $D = 0$
$[-(7p + 2)^2] - 4(2p + 1)(7p - 3) = 0 \implies 7p^2 - 24p - 16 = 0 \implies (7p + 4)(p - 4) = 0 \implies p = 4, p = -\frac{4}{7}$
For $p = 4$, the equation is $9x^2 - 30x + 25 = 0$ whose roots are $\frac{5}{3}, \frac{5}{3}$
For $p = -\frac{4}{7}$, the equation is $x^2 - 14x + 49 = 0$ whose roots are $7, 7$
$[-(7p + 2)^2] - 4(2p + 1)(7p - 3) = 0 \implies 7p^2 - 24p - 16 = 0 \implies (7p + 4)(p - 4) = 0 \implies p = 4, p = -\frac{4}{7}$
For $p = 4$, the equation is $9x^2 - 30x + 25 = 0$ whose roots are $\frac{5}{3}, \frac{5}{3}$
For $p = -\frac{4}{7}$, the equation is $x^2 - 14x + 49 = 0$ whose roots are $7, 7$