Question

if one of the roots of the quadratic equation x2+6x+p=0 is twice the other root, find the value of p

Answer 1

the roots are a &2a
then (x-a)(x-2a)=0
x2 -3ax +2a^2 =0
by comparing with the given equation
-3a=6
then a=-2
p=2a^2 = 8

Answer 2

x2+6x+p=0 has r = c , 2c
c^2 +6c +p = 0
4c^2 + 12c + p = 0
=> 3c^2 + 6c = 0
=> 3c (c +2) = 0
=> c = 0 and c = -2
c = -2 => 4 -12 = -p
Thus p = 8