1. Oct 12, 2014

### LiHJ

1. The problem statement, all variables and given/known data
Dear Mentors and Helpers,

Here's the question:
Find the possible values of k such that one root of the equation 2x^2 + kx + 9 = 0 is twice the other.

2. Relevant equations
My classmate's working:

Discriminate > 0
k^2 - (4)(2)(9) > 0
k^2 -72 > 0
[k + sqrt (72)] [k- sqrt(72)] > 0

Answer: k > sqrt (72) or k < - sqrt(72)

3. The attempt at a solution
My working:

Let p and 2p be the roots of the equation.

Sum of roots:
3p = (-k)/2
p = (-k)/6 -----(1)

Product of roots:
2p^2 = 9/2 ------(2)

Substitute (1) into (2):

2(-k/6)^2 = 9/2
(k^2)/36 = 9/4
k^2 = 81
k = +/- (9)

Dear Mentors and Helpers,

Last edited by a moderator: Oct 12, 2014
2. Oct 12, 2014

### Staff: Mentor

I get the same thing you did. Your classmate's work is incomplete. All he did was to find intervals of k values for which the discriminant is positive.

You can take this problem one step further to verify that your work is correct, by finding the two roots of the quadratic. One of the roots should be twice the other. Note that there are two pairs of values that work.

Also, I approached this problem in a different way, since I don't have the sum of roots, product of roots formulas memorized.

I rewrote the equation as x2 + (k/2)x + 9/2 = 0. Then, since p and 2p are roots, it must be that (x - p)(x - 2p) = 0. Expand the second equation and then equate the coefficient of x and the constant term in the two equations. This will give you two equations in the unknowns k and p.

3. Oct 12, 2014

### LiHJ

Thanks for the verification and further explanation of another method