Find the values of p for which the equadratic equation (2p + 1) $x^{2}$ - (7p + 2) x + (7p - 3) = 0 has equal roots. Also find these roots.

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Solution

Given Eqn is $\frac{(2 p + 1) x^{2}}{a} - \frac{(7 p + 2) x}{b} + \frac{7 p - 3}{c} = 02$
If this eqn has equal roots Discriminant
$D = 0$
$b^{2} - 4 a c = 0$
$\Rightarrow (7 p + 2)^{2} - 4 (2 p + 1) (7 p - 3) = 0$
$\Rightarrow 49 p^{2} + 4 + 28 p - 4 (14 9^{2} - 6 p - 3 + 7 p) = 0$
$\Rightarrow 49 p^{2} + 4 + 28 p - (56 p^{2} - 24 p - 12 + 28 p) = 0$
$\Rightarrow 49 p^{2} + 4 + 28 p - 56 p^{2} - 4 p + 12 = 0$
$\Rightarrow - 7 p^{2} + 24 p + 16 = 0$
$\Rightarrow 7 p^{2} - 24 p - 16 = 0 \Rightarrow P = 4, \frac{- 4}{7}$
Now Roots At
$i) P = 4$ $i i) P = \frac{- 4}{7}$
Eqn is $9 x^{2} - 30 x + 25 = 0$ Eqn is $\frac{- 1}{7} x^{2} + 2 x - 7 = 0$
$3 x - 5 = 0$ $x^{2} - 14 x + 49 = 0$
$x = \frac{5}{3}$ $x = 7$

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