Roots and coeffecients of polynomial equations

In summary, you solved the equation2x2-3\beta+\alpha\beta=0by dividing it by 2 and solving for x and k.
  • #1
olee
24
0

Homework Statement


One root of 2x2-kx+k=0 is twice the other. Find K. (assume k is not 0)


Homework Equations


not sure what this means.


The Attempt at a Solution



[tex]\alpha[/tex]+[tex]\beta[/tex]=[tex]\frac{-b}{a}[/tex]

[tex]\alpha\beta[/tex]=[tex]\frac{c}{a}[/tex]

then i tried

[tex]\alpha[/tex]=2[tex]\beta[/tex] then substituted in..

2x2-3[tex]\beta[/tex]+[tex]\alpha\beta[/tex]=0 and now I am stuck

im not sure if this is right way to do this
 
Physics news on Phys.org
  • #2
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
Mark44 said:
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.

thanks for reply. which equation are you mentioning?
 
  • #4
The only equation you were given, and the one whose roots you are trying to find.
2x2 - kx + k=0
 
  • #5
Mark44 said:
The only equation you were given, and the one whose roots you are trying to find.
2x2 - kx + k=0

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
You should have two equations in the two unknowns. Solve these equation simultaneously to find r and k.
 
  • #7
Mark44 said:
You should have two equations in the two unknowns. Solve these equation simultaneously to find r and k.

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 = -[tex]\frac{b}{2}[/tex])

and then where the +k is (which is c) i substitute in 4r2? (to get this i rearranged 2r2=[tex]\frac{c}{2}[/tex]
 
  • #8
olee said:
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 = -[tex]\frac{b}{2}[/tex] )

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

edit: was to remote the extra bracket
 
  • #9
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
You got the equation

2x2-3[tex]\beta[/tex]+[tex]\alpha\beta[/tex]=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[tex]\beta[/tex] x + [tex]\alpha\beta[/tex]=0
 
Last edited:
  • #11
Mark44 said:
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.

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




epenguin said:
You got the equation

2x2-3[tex]\beta[/tex]+[tex]\alpha\beta[/tex]=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[tex]\beta[/tex] x + [tex]\alpha\beta[/tex]=0


thanks for reply. this method i ended up with 2x2-3[tex]\beta[/tex]x+2[tex]\beta[/tex]2
 
  • #12
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
Mark44 said:
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?

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
olee said:
thanks for reply. doing this way i made the two equations equal each other (2r2-kr+k = 8r2-2kr+k
ended up with r=[tex]\frac{6}{k}[/tex] and i can't solve that because of two unknown variables.

thanks for reply. this method i ended up with 2x2-3[tex]\beta[/tex]x+2[tex]\beta[/tex]2

Sorry -I made a typo.

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[tex]\beta[/tex]x+2[tex]\beta[/tex]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:
  • #15
epenguin said:
Sorry -I made a typo.

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[tex]\beta[/tex]x+2[tex]\beta[/tex]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.

so how do i solve it from here? I am really confused..
 
  • #16
Any, in other words every, quadratic with one root twice the other is of form

[tex]x^2-3\beta x+2\beta^2[/tex] (1)

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

You are told

[tex]x^2 - \frac{kx}{2} + \frac{k}{2} [/tex] (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:

What are roots and coefficients of polynomial equations?

Roots and coefficients are important components of polynomial equations. The roots are the values of x that make the equation equal to 0, while the coefficients are the numerical values that are multiplied by the variable terms.

How do I find the roots of a polynomial equation?

To find the roots of a polynomial equation, you can use a variety of methods such as the quadratic formula, factoring, or graphing. These methods can help you determine the values of x that make the equation equal to 0.

What is the relationship between the number of roots and the degree of a polynomial equation?

The degree of a polynomial equation is equal to the highest exponent of the variable term. The number of roots of a polynomial equation is equal to the degree of the equation. This means that a polynomial equation of degree n will have n roots.

Can a polynomial equation have more than one root?

Yes, a polynomial equation can have multiple roots. This is especially true for equations with higher degrees. For example, a cubic equation with degree 3 can have up to 3 roots.

How do I determine the coefficients of a polynomial equation?

The coefficients of a polynomial equation can be determined by looking at the numerical values that are multiplied by the variable terms. For example, in the equation 2x^2 + 3x + 1, the coefficients are 2, 3, and 1.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
20
Views
1K
  • Precalculus Mathematics Homework Help
Replies
10
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Precalculus Mathematics Homework Help
Replies
21
Views
3K
  • Precalculus Mathematics Homework Help
Replies
15
Views
3K
  • Precalculus Mathematics Homework Help
Replies
6
Views
651
Back
Top