If the quadratic equation 6x^2 + nx + 18 = 0 has one root twice that of the other what is n?

The roots of the quadratic equation ax^2 + bx + c = 0 are
given by [-b + sqrt (b^2 - 4ac)]/2a and [-b - sqrt (b^2 -
4ac)]/2a.

If one root of 6x^2 + nx + 18 = 0 is twice the
other root the ratio [-b + sqrt (b^2 - 4ac)]/2a / [-b - sqrt (b^2 - 4ac)]/2a should be 2
or 0.5

[[-b + sqrt (b^2 - 4ac)]/2a] / [[-b - sqrt (b^2 -
4ac)]/2a]

=> [-b + sqrt (b^2 - 4ac)] / [-b - sqrt
(b^2 - 4ac)]

[-n + sqrt (n^2 - 18*6*4)] / [ -n - sqrt(n^2 -
4*6*18] = 2

=> -n + sqrt(n^2 - 4*6*18) = -2n -
2*sqrt(n^2 - 4*6*18)

=> n = -3*sqrt(n^2 -
4*6*18)

=> n^2 = 9*(n^2 -
4*6*18)

=> 8n^2 =
9*4*6*18

=> n^2 =
486

=> n = sqrt 486 or n = -sqrt
486

[-n + sqrt (n^2 - 18*6*4)] / [ -n - sqrt(n^2 - 4*6*18]
= 1/2

=> -2n + 2*sqrt(n^2 - 4*6*18) = -n - sqrt(n^2
- 4*6*18)

=> n = 3*sqrt(n^2 -
4*6*18)

=> n^2 = 9*(n^2 -
4*6*18)

=> n^2 =
486

=> n = sqrt 486 and n = -sqrt
486

The required values of n are sqrt 486 and
n = -sqrt 486