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