If 3 is a root of the quadratic equation $x^{2} - x + k = 0$ , find the value of $p$ so that the roots of the equation $x^{2} + k (2 x + k + 2) + p = 0$ are equal.

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Solution

Given: $3$ is a root of equation $x^{2} - x + k = 0$

Substitute the value of $x = 3$

$(3)^{2} - (3) + k = 0$
$k = - 6$

Now, $x^{2} + k (2 x - 6 + 2) + p = 0$
$x^{2} + (- 6) (2 x - 6 + 2) + p = 0$
$x^{2} - 12 x + 36 - 12 + p = 0$
$x^{2} - 12 x + (24 + p) = 0$

Compare given equation with the general from of quadratic equation, which $a x^{2} + b x + c = 0$
$a = 1, b = - 12, c = 24 + p$

Find Discriminant:
$D = b^{2} - 4 a c$
$= (- 12)^{2} - 4 \times 1 \times (24 + p)$
$= 144 - 96 - 4 p = 48 - 4 p$

Since roots are real and equal, put $D = 0$
$48 - 4 p = 0$
$4 p = 48$
$p = 12$

The value of $p$ is $12$

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