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.
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~ :)
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?
I had not considered the possibility of complex coefficients in the equation. Nice work by Jai.
Back to physics for me
To find the possible values of k, let's first express the roots of the quadratic equation in terms of a variable, say "r".
Let one root of the quadratic equation be r. Since it is given that one root is one-third the other root, the other root can be expressed as 3r.
Now, we can write the quadratic equation using the given roots:
(x - r)(x - 3r) = 0
Expanding this equation gives us:
x^2 - (r + 3r)x + 3r^2 = 0
Simplifying the equation further:
x^2 - 4rx + 3r^2 = 0
Comparing this equation with x^2 + kx + 12 = 0, we can see that:
k = -4r
12 = 3r^2
From the second equation, we get:
r^2 = 4
r = ±2
Now, substitute the value of r into the expression for k:
k = -4r
k = -4(±2)
k = -8 or k = 8
Therefore, the possible values of k are -8 and 8.