SOLUTION: The equation 2(x^2+12)+kx=0 where k <0 has roots a and b where a-b=4.Find the value of k

Question 457996: The equation 2(x^2+12)+kx=0
where k <0 has roots a and b where a-b=4.Find the value of k
Answer by J2R2R(94) (Show Source): You can put this solution on YOUR website!
2x^2 + kx + 24 = 0
Two roots a and b where a - b = 4 so a must be the larger.

We know that
a = (-k + sqrt(k^2 - 192))/4
b = (-k - sqrt(k^2 - 192))/4

Therefore a - b = 2 (sqrt(k^2 - 192))/4 = 4
sqrt(k^2 - 192) = 8
k^2 - 192 = 64
k^2 = 192 + 64
k^2 = 256
k = -16 (k<0 so cannot be 16)

Check
2x^2 - 16x + 24 = 0
x^2 - 8x + 12 = 0
(x - 6)(x - 2) = 0
a = 6; b = 2
a - b = 4

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