# Homework Help: Roots and coeffecients of polynomial equations

1. Mar 2, 2010

### olee

1. The problem statement, all variables and given/known data
One root of 2x2-kx+k=0 is twice the other. Find K. (assume k is not 0)

2. Relevant equations
not sure what this means.

3. The attempt at a solution

$$\alpha$$+$$\beta$$=$$\frac{-b}{a}$$

$$\alpha\beta$$=$$\frac{c}{a}$$

then i tried

$$\alpha$$=2$$\beta$$ then substituted in..

2x2-3$$\beta$$+$$\alpha\beta$$=0 and now im stuck

im not sure if this is right way to do this

2. Mar 2, 2010

### Staff: Mentor

Let r and 2r be the two roots. Substitute each one in your equation. That will give you two equations in two unknowns - r and k. Solve the two equations simulataneously for these variables.

3. Mar 2, 2010

### olee

thanks for reply. which equation are you mentioning?

4. Mar 2, 2010

### Staff: Mentor

The only equation you were given, and the one whose roots you are trying to find.
2x2 - kx + k=0

5. Mar 2, 2010

### olee

thanks for reply. but i still have two unknowns and i can't solve

edit: and i still have the x at the start of that equation. all i know is 'a' (which is 2)

6. Mar 2, 2010

### Staff: Mentor

You should have two equations in the two unknowns. Solve these equation simultaneously to find r and k.

7. Mar 2, 2010

### olee

by substituting in, do you mean where the -k is (which is b) i substitute in 2r+4r? (to get this i had r+2r = -$$\frac{b}{2}$$)

and then where the +k is (which is c) i substitute in 4r2? (to get this i rearranged 2r2=$$\frac{c}{2}$$

8. Mar 2, 2010

### olee

edit: was to remote the extra bracket

9. Mar 2, 2010

### Staff: Mentor

What I'm saying is that you should ignore the equations you have with alpha and beta and a and b, and focus on the equation you're trying to solve, 2x2 - kx + k = 0.

You are given that r and 2r are roots of this equation, so if you replace x by r you'll get one equation, and replace x by 2r, you'll get another equation. These two equations will be in terms of r and k.

Solve those two equations simultaneously for r and k.

10. Mar 2, 2010

### epenguin

You got the equation

2x2-3$$\beta$$+$$\alpha\beta$$=0

In the 2nd term here you have used the information you are given (one root is twice the other) and in the third term you haven't yet.

Oh, and you must have meant

2x2- 3$$\beta$$ x + $$\alpha\beta$$=0

Last edited: Mar 2, 2010
11. Mar 2, 2010

### olee

thanks for reply. doing this way i made the two equations equal eachother (2r2-kr+k = 8r2-2kr+k
ended up with r=$$\frac{6}{k}$$ and i can't solve that because of two unknown variables.

thanks for reply. this method i ended up with 2x2-3$$\beta$$x+2$$\beta$$2

12. Mar 2, 2010

### Staff: Mentor

I get r = k/6 (not 6/k) or r = 0.

If r = 0, then 2r = 0, so what is k?
If r = k/6, then 2r = k/3, so what is k?

13. Mar 2, 2010

### olee

thanks for reply. yeah i got r=k/6 (i mistyped)

If r = 0, then 2r = 0, so what is k? k=0
If r = k/6, then 2r = k/3, so what is k? plugged r=k/6 back into original equation and got k = 9 or 0.. right?

EDIT: made a mistake, recalculated and got 9 and 0

14. Mar 2, 2010

### epenguin

It is usually less confusing to work with an equation whose leading coefficient is 1. So better to divide your initial equation by 2. So use

x2 - kx/2 + k/2

A quadratic with one root twice the other is

x2-3$$\beta$$x+2$$\beta$$2

I think you had the idea of that but the leading coefficient caused you confusion. You should easily be able to see your last result must be wrong; do not accept any reult without checking it against your original problem.

Last edited: Mar 2, 2010
15. Mar 2, 2010

### olee

so how do i solve it from here? im really confused..

16. Mar 2, 2010

### epenguin

Any, in other words every, quadratic with one root twice the other is of form

$$x^2-3\beta x+2\beta^2$$ (1)

(a result you got but for a little mistake.)

You are told

$$x^2 - \frac{kx}{2} + \frac{k}{2}$$ (2)

is a quadratic with one root twice the other.

So (2) must be of form (1). Or if you like identical to it for all x. So do you see how to compare them (remembering similar exercises you've probably done before for identical expressions)?

Last edited: Mar 2, 2010