90
Find the value(s) of $p$ for which the quadratic equation given as $(p + 4)x^2 - (p + 1)x + 1 = 0$ has real and equal roots. Also, find the roots of the equation(s) so obtained.
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For real and equal roots, $D = 0$
$\therefore [-(p + 1)]^2 - 4(p + 4) = 0$
$\implies p^2 - 2p - 15 = 0$
$\implies (p - 5)(p + 3) = 0$
$\therefore p = 5, -3$
For $p = 5, 9x^2 - 6x + 1 = 0 \implies (3x - 1)(3x - 1) = 0 \therefore x = \frac{1}{3}, \frac{1}{3}$
For $p = -3, x^2 + 2x + 1 = 0 \implies (x + 1)(x + 1) = 0 \therefore x = -1, -1$
Hence roots are $\frac{1}{3}, \frac{1}{3}$ and $-1, -1$ for $p = 5$ and $p = -3$ respectively.
$\therefore [-(p + 1)]^2 - 4(p + 4) = 0$
$\implies p^2 - 2p - 15 = 0$
$\implies (p - 5)(p + 3) = 0$
$\therefore p = 5, -3$
For $p = 5, 9x^2 - 6x + 1 = 0 \implies (3x - 1)(3x - 1) = 0 \therefore x = \frac{1}{3}, \frac{1}{3}$
For $p = -3, x^2 + 2x + 1 = 0 \implies (x + 1)(x + 1) = 0 \therefore x = -1, -1$
Hence roots are $\frac{1}{3}, \frac{1}{3}$ and $-1, -1$ for $p = 5$ and $p = -3$ respectively.