Question

if 3 is the root of the quadratic equation X square minus x + K is equal to zero find the value of P so that the roots of the equation X square + K ( 2 X + K + 2) + p is equal to zero are equal

Answer 1

Putting X= 3

x^2 - x + K = 0

(3)^2 -3 + k = 0

9-3 + k = 0

6 + k = 0

k = -6

for 2nd polynomial, putting k= -6

x^2 + (-6)(2x-6+2) + p = 0

x^2 -6(2x-4) + p = 0

x^2 -12x + 24 +p = 0

for equal roots D= 0

B square - 4AC = 0

(-12) square - 4 (1)(24+p) = 0

144 = 4p + 96

144-96 = 4p

48 = 4p

p= 12

Answer 2

Answer:

12

Step-by-step explanation:

x^2-x+k=0

Since 3 is root of given equation so putting x=3,

(3)^2-3+k=0

k=-6

x^2+k(2x+k+2)+p=0

Putting k=-6 in above equation,

x^2+2*(-6)x+(-6)^2+2*(-6)+p=0

x^2-12x-24+p=0

Since root are equal therefore D=0

b^2-4ac=0

(-12)^2-4*1*(-24+p)=0

p= 12