Asked by T

Given one of the roots of the quadratic equation x^2+kx=12 is one third the other root. Find the possible values of k.

Answers

Answered by Jai
let a = one root
let (1/3)a = the other root
recall that quadratic equation can also be written as,
x^2 - (sum of roots)x + (product of products) = 0

thus,
product of roots = -12
-12 = (1/3)a^2
-36 = a^2
a = 6i
(1/3)a = 2i
*note that i = sqrt(-1) , which is an imaginary number

sum of roots = -k
-k = 6i + 2i
k = -8i

hope this helps~ :)
Answered by drwls
It seems to me the product of the two roots must be -12, which requires that one has to be positive and the other negative NO MATTER WHAT THE VALE OF k IS. I don't see how the ratio of the roots can then be +3. Do you mean the ratio of the absolute values of the roots must be three?
Answered by drwls
I had not considered the possibility of complex coefficients in the equation. Nice work by Jai.

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