Find the value(s) of p for which the quadratic equation (2p + 1)x² - (7p + 2)x + (7p - 3) = 0 has equal roots. Also find these roots.

The given equation is (2p + 1)x² - (7P + 2)x + (7p - 3) = 0.

Comparing with ax² + bx + c = we obtain,
a = 2p + 1 , b = -(7p + 2) , c = 7p - 3

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (-(7p + 2))^2 - 4 \times (2p + 1) \times (7p - 3) \\[1em] = 49p^2 + 4 + 28p - 4(2p + 1)(7p - 3) \\[1em] = 49p^2 + 4 + 28p - 4(14p^2 - 6p + 7p - 3) \\[1em] = 49p^2 - 56p^2 + 28p - 4p + 4 + 12 \\[1em] = -7p^2 + 24p + 16$

For equal roots, discriminant = 0

$\Rightarrow -7p^2 + 24p + 16 \\[1em] \Rightarrow 7p^2 - 24p - 16 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow 7p^2 - 28p + 4p - 16 = 0 \\[1em] \Rightarrow 7p(p - 4) + 4(p - 4) = 0 \\[1em] \Rightarrow (7p + 4)(p - 4) = 0 \\[1em] \Rightarrow 7p + 4 = 0 \text{ or } p - 4 = 0 \\[1em] \Rightarrow 7p = -4 \text{ or } p = 4 \\[1em] \Rightarrow p = -\dfrac{4}{7} \text{ or } p = 4 \\[1em]$

When p = $-\dfrac{4}{7}$ the equation becomes $(2 \times -\dfrac{4}{7} + 1)x^2 - (7 \times -\dfrac{4}{7} + 2)x + (7 \times -\dfrac{4}{7} - 3) = 0$ so the roots are :

$\Rightarrow -\dfrac{1}{7}x^2 + 2x - 7 = 0 \\[1em] \Rightarrow x^2 - 14x + 49 = 0 \text{ (On multiplying complete equation by -7) } \\[1em] \Rightarrow x^2 - 14x + 49 = 0 \\[1em] \Rightarrow x^2 - 7x - 7x + 49 = 0 \\[1em] \Rightarrow x(x - 7) - 7(x - 7) = 0 \\[1em] \Rightarrow (x - 7)(x - 7) = 0 \\[1em] \Rightarrow x - 7 = 0 \text{ or } x - 7 = 0 \\[1em] \Rightarrow x = 7 \text{ or } x = 7 \\[1em]$

When p = 4 the equation becomes $(2 \times 4 + 1)x^2 - (7 \times 4 + 2)x + (7 \times 4 - 3) = 0$ so the roots are :

$\Rightarrow 9x^2 - 30x + 25 = 0 \\[1em] \Rightarrow 9x^2 - 15x - 15x + 25 = 0 \\[1em] \Rightarrow 3x(3x - 5) - 5(3x - 5) = 0 \\[1em] \Rightarrow (3x - 5)(3x - 5) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } 3x - 5 = 0 \\[1em] \Rightarrow x = \dfrac{5}{3} \text{ or } x = \dfrac{5}{3} \\[1em]$

(Ans.) 4, $-\dfrac{4}{7}$ ; when p = $-\dfrac{4}{7}$, roots are 7, 7 and when p = 4, roots are $\dfrac{5}{3} , \dfrac{5}{3}.$