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Math Help - Roots of quadratic equations.

  1. #1
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    Roots of quadratic equations.

    I've made it to the last question on a homework piece, but I just can't figure out how to start. The question is -

    One root of the equation x - px + q = 0 is twice the other root. Show that 2p = 9q.

    Originally, I just re-arranged 2p=9q to p=(sqrt)9q/2, and subbed in some numbers, until I got p=4.5, q=9. But my maths tutor said that's not the answer he's looking for, and that I should be using alpha and beta, and to try again tonight. I've spent the last 45 mins scribbling bits out to no avail, I just don't know how to get the values of alpha and beta to prove 2p=9q.

    Any suggestions?
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  2. #2
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    Quote Originally Posted by Mbrown141 View Post
    I've made it to the last question on a homework piece, but I just can't figure out how to start. The question is -

    One root of the equation x - px + q = 0 is twice the other root. Show that 2p = 9q.

    Originally, I just re-arranged 2p=9q to p=(sqrt)9q/2, and subbed in some numbers, until I got p=4.5, q=9. But my maths tutor said that's not the answer he's looking for, and that I should be using alpha and beta, and to try again tonight. I've spent the last 45 mins scribbling bits out to no avail, I just don't know how to get the values of alpha and beta to prove 2p=9q.

    Any suggestions?
    let the two roots be a and b and you are told that a=2b -- 1

    sum of roots, a+b = p -- 2

    product of roots, ab = q -- 3

    You should be able to take it from here. Just think about what you can do with the 3 simultaneous equations.
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  3. #3
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    Quote Originally Posted by Mbrown141 View Post
    I've made it to the last question on a homework piece, but I just can't figure out how to start. The question is -

    One root of the equation x - px + q = 0 is twice the other root. Show that 2p = 9q.

    Originally, I just re-arranged 2p=9q to p=(sqrt)9q/2, and subbed in some numbers, until I got p=4.5, q=9. But my maths tutor said that's not the answer he's looking for, and that I should be using alpha and beta, and to try again tonight. I've spent the last 45 mins scribbling bits out to no avail, I just don't know how to get the values of alpha and beta to prove 2p=9q.

    Any suggestions?
    (x - \alpha)(x - 2 \alpha) = x^2 - 3 \alpha x + 2 \alpha^2.

    Therefore:

    \displaystyle p = 3 \alpha \Rightarrow \frac{p}{3} = \alpha .... (1)

    \displaystyle q = 2 \alpha^2 \Rightarrow \frac{q}{2} = \alpha^2 .... (2)

    Square equation (1) and equate the result to equation (2).
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    let the two roots be a and b and you are told that a=2b -- 1

    sum of roots, a+b = p -- 2

    product of roots, ab = q -- 3

    You should be able to take it from here. Just think about what you can do with the 3 simultaneous equations.
    Already had that, my problem is not having any clue where to take it. See, I thought -

    p=a+b
    2p=2(a+b)
    2p=(2(a+b))

    But how do I relate that to q? Q = ab, which doesn't seem to help here. Am I taking this the complete wrong way?

    Quote Originally Posted by mr fantastic View Post
    (x - \alpha)(x - 2 \alpha) = x^2 - 3 \alpha x + 2 \alpha^2.

    Therefore:

    \displaystyle p = 3 \alpha \Rightarrow \frac{p}{3} = \alpha .... (1)

    \displaystyle q = 2 \alpha^2 \Rightarrow \frac{q}{2} = \alpha^2 .... (2)

    Square equation (1) and equate the result to equation (2).
    I understand all that so far, thanks for that. Next step is to square a part of the first equation/result?

    If p=3 \alpha, then 2p=6 \alpha^2.

    If q= \frac{\alpha^2}{2}, then 9q=4.5 \alpha^2, which isn't right. Not sure where I've gone wrong.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    (x - \alpha)(x - 2 \alpha) = x^2 - 3 \alpha x + 2 \alpha^2.

    Therefore:

    \displaystyle p = 3 \alpha \Rightarrow \frac{p}{3} = \alpha .... (1)

    \displaystyle q = 2 \alpha^2 \Rightarrow \frac{q}{2} = \alpha^2 .... (2)

    Square equation (1) and equate the result to equation (2).
    I understand all that so far, thanks for that. Next step is to square a part of the first equation/result?

    If p=3 \alpha, then 2p=6 \alpha^2.

    If q= \frac{\alpha^2}{2}, then 9q=4.5 \alpha^2, which isn't right. Not sure where I've gone wrong.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    \dfrac{p}{3} = \alpha

    When you square this you get:

    \dfrac{p^2}{9} = \alpha^2

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  7. #7
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    Quote Originally Posted by Unknown008 View Post
    \dfrac{p}{3} = \alpha

    When you square this you get:

    \dfrac{p^2}{9} = \alpha^2

    So you're saying because squaring \dfrac{p}{3} = \alpha = \dfrac{p^2}{9} = \alpha^2, then multiplying both sides by 9 gives p^2 = 9\alpha^2, but since beta is twice alpha, I need to double the p?

    That would give the correct answer, just checking my understanding is right?

    Thanks.

    Edit: This leaves the wrong answer somehow. 2p=9q, whereas I end up with 2p=9q...
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  8. #8
    MHF Contributor Unknown008's Avatar
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    If you look at what Mr. Fantastic gave you, he didn't include any beta. He used directly twice alpha instead of having alpha and beta.

    If he used beta, then you'd have:

        (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta) x + \alpha\beta

    Then;

    p = \alpha + \beta

    q = \alpha\beta

    Then, you know that \beta = 2\alpha

    This gives:

    p = \alpha + 2\alpha = 3 \alpha

    q = \alpha(2\alpha) = 2\alpha^2

    Which comes back to what Mr. Fantastic gave you
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  9. #9
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    Quote Originally Posted by Unknown008 View Post
    If you look at what Mr. Fantastic gave you, he didn't include any beta. He used directly twice alpha instead of having alpha and beta.

    If he used beta, then you'd have:

        (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta) x + \alpha\beta

    Then;

    p = \alpha + \beta

    q = \alpha\beta

    Then, you know that \beta = 2\alpha

    This gives:

    p = \alpha + 2\alpha = 3 \alpha

    q = \alpha(2\alpha) = 2\alpha^2

    Which comes back to what Mr. Fantastic gave you
    Thanks for explaining that. I understand up to that point, so I'm going to use that in the homework piece, and I know some marks are awarded for getting that far. I have no clue how to get from there to 2p=9q, but some marks are better than none!

    Thanks.
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  10. #10
    MHF Contributor Unknown008's Avatar
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    It's easy from there on. You can substitute the alpha in second equation, or equate alpha squared.

    Like if you had

    p + q = r
    pq = 2r

    Using the first equation and the second equation, I can express p and q without r.

    pq = 2(p + q)

    It's nearly the same thing in your problem.
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  11. #11
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    Quote Originally Posted by Unknown008 View Post
    It's easy from there on. You can substitute the alpha in second equation, or equate alpha squared.

    Like if you had

    p + q = r
    pq = 2r

    Using the first equation and the second equation, I can express p and q without r.

    pq = 2(p + q)

    It's nearly the same thing in your problem.
    So you're saying using the

    \alpha=\frac{p}{3}, put that into the second equation in place of alpha, resulting in q=\frac{2p}{3}. Multiplying both sides by 3, gets 3q=2p, when it needs to be 9q=2p...

    Or, do I square both top and bottom of the fraction? So I'd have q=2(( \frac{p}{3}))? That would give q=2\frac{p^2}{9}, then multiplying both sides by 9 would get 2p=9q. That sounds right, anyone back me up?
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  12. #12
    MHF Contributor Unknown008's Avatar
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    Yep, that's it

    You square both the numerator and the denominator. The square is for both of them.
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