Question

If 3 is a root of the quadratic equation x²-x+k=0, find the value of p so that the roots of the equation x²+k(2x+k+2)+p=0 are equal.


Solve it ASAP ^_^​

Answer 1

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

Compare given equation with the general from of quadratic equation, which ${ax}^{2} + bx + c = 0$

$⇒a = 1 ,b = - 12, c = 24 + p$

Find Discriminant:

$D = {b}^{2} - 4ac \\ = { - 12}^{2} - 4 \times 1 \times (24 + p) \\ = 144 - 96 - 4p \\ = 48 - 4p$

Since roots are real and equal, put $D=0$

$⇒48 - 4p = 0 \\ ⇒4p = 48 \\ ⇒p = \frac{48}{4} \\⇒p = 12$