If 3 is 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

substitute x=3 in eqn. $x^{2} - x + k = 0 (3)^{2} - 3 + k = 0 9 - 3 + k = 0 6 + k = 0 k = - 6$

Now substitute k value in equation

$x^{2} + k (2 k + k + 2) + p = 0$

we get

$x^{2} - 12 x + 24 + p = 0 a = 1, b = - 12, c = 24 + p S i n c e r o o t s a r e e q u a l, b^{2} - 4 a c = 0 = (- 12)^{2} - 4 (1) (24 + p) = 144 - 96 - 4 p = 0 48 - 4 p = 0 - 4 p = - 48 p = 12$

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