Question

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( 2x + k + 2 ) + p = 0 are equal.

Answer 1

here is the answer...the answer is 12..

Answer 2

Heya mate, Here is ur answer

P(x) = x^2 -x +k

Given ; x = 3

P(3) = 3^2 -3+k

0= 9 -3 + k

0= 6+ k

-6 = K

So,

x^2 +k (2x+k+2)+p

=x^2 + (-6)(2x+(-6) +2)+p

= x^2-(6)(2x-6+2) +p

=x^2 -(6)(2x-4) +p

=x^2 -12x +24 + p

a =1 , b=-12 c = 24+p

D= 0 (as roots are equal)

${b}^{2} - 4ac = 0$


$( - 12) {}^{2} - 4 \times 1 \times (24 + p) = 0$


$144 - 4(24 + p) = 0$


$144 - 96 - 4p = 0$

$48 - 4p = 0$


$- 4p = - 48$


$p = \frac{ - 48}{ - 4}$


$p = 12$

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@Laughterqueen

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