Question

one of the roots of the equation x^2 - 2qx + 2q +4 = 0 is three times the other root, determine the value q

Answer 1

one root
(2q+sqrt(4q^2-8q-16))/2
other root
(2q -sqrt(4q^2-8q-16))/2

but one root=3*other root or
(2q+sqrt(4q^2-8q-16)=6q-3sqrt(4q^2-8q-16)

8q=-4sqrt( )
divide by 4, square both sides..
4q^2=4q^2-8q-16)
q=(-2)

check my work. Typing is hard to do without making a mistake.

Answer 2

recall that for ax^2 + bx + c = 0 ,
the sum of the roots = -b/a
product of the roots = c/a


we are given that the roots are r and 3r
so from your equation of
x^2 - 2qx + 2q+4 = 0
a = 1, b = -2q, c = 2q+4

r+3r = 2q/1
4r = 2q
q = 2r

r(3r) = 2q+4
3r^2 = 2(2r) + 4
3r^2 - 4r - 4 = 0
(r-2)(3r +2) = 0
r = 2 or r = -2/3

<b>if r = 2, q = 4
ir r = -2/3) , q = -4/3</b>

check for q=4, then the equation is
x^2 - 8x + 12 = 0
(x-2)(x-6) = 0
x = 2 or x = 6
and one root is 3 times the other.
checking for q = -4/3 is a bit more messy, but still varifies.