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Thread: Quadratic Equations

  1. #1
    Junior Member
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    Quadratic Equations

    Hi,
    So here's my problem:
    If the equation ax^2 + bx + c = 0 has roots alpha and beta, find the roots of the equation:
    ax^2 - bx(x-1) + c(x-1)^2 = 0

    What I did:
    I managed to simplify the given equation to:
    x^2(a-b+c) + x(b - 2c) + c = 0

    And now I don't know how to proceed.
    A slight push in the right direction would help.

    Cheers.
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  2. #2
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    Quote Originally Posted by Ruler of Hell View Post
    Hi,


    What I did:
    I managed to simplify the given equation to:
    x^2(a-b+c) + x(b - 2c) + c = 0
    Cheers.
    Havent checked you simplification but if it is correct then you problem is simple just use the quadratic formula $\displaystyle ax^2 +bx +c = 0$
    you a is (a+b+c)
    you b is (b-2c)
    Can you take it from here? or more pushing needed? =)
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  3. #3
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    Quote Originally Posted by Ruler of Hell View Post
    Hi,
    So here's my problem:
    If the equation ax^2 + bx + c = 0 has roots alpha and beta, find the roots of the equation:
    ax^2 - bx(x-1) + c(x-1)^2 = 0

    What I did:
    I managed to simplify the given equation to:
    x^2(a-b+c) + x(b - 2c) + c = 0

    And now I don't know how to proceed.
    A slight push in the right direction would help.

    Cheers.
    If $\displaystyle \alpha$ and $\displaystyle \beta$ are roots of $\displaystyle ax^2+ bx+ c= 0$ then you know that $\displaystyle ax^2+ bx+ c= a(x- \alpha)(x- \beta)= ax^2- a(\alpha+ \beta)x+ a\alpha\beta$.

    That tells you that $\displaystyle b= a(\alpha+ \beta)$ and $\displaystyle c= \alpha\beta$.
    So what are a- b+ c and b- 2c in terms of a, $\displaystyle \alpha$, and $\displaystyle \beta$?

    Once you know that, you should be able to use that to find roots of the second equation in terms of a, $\displaystyle \alpha$, and $\displaystyle \beta$.
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Ruler of Hell View Post
    Hi,
    So here's my problem:
    If the equation ax^2 + bx + c = 0 has roots alpha and beta, find the roots of the equation:
    ax^2 - bx(x-1) + c(x-1)^2 = 0

    What I did:
    I managed to simplify the given equation to:
    x^2(a-b+c) + x(b - 2c) + c = 0

    And now I don't know how to proceed.
    A slight push in the right direction would help.

    Cheers.
    $\displaystyle (x-\alpha)(x - \beta) = (a-b+c)x^2 + (b-2c)x + c$

    This means that

    $\displaystyle a-b+c = 1$

    $\displaystyle \alpha \beta = c$

    $\displaystyle \alpha + \beta = (b-2c)$


    Alternatively use the quadratic formula
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  5. #5
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    Quote Originally Posted by Ruler of Hell View Post
    Hi,
    So here's my problem:
    If the equation ax^2 + bx + c = 0 has roots alpha and beta, find the roots of the equation:
    ax^2 - bx(x-1) + c(x-1)^2 = 0

    What I did:
    I managed to simplify the given equation to:
    x^2(a-b+c) + x(b - 2c) + c = 0

    And now I don't know how to proceed.
    A slight push in the right direction would help.

    Cheers.
    As $\displaystyle a,\ b$ and $\displaystyle c$ are constants, along with $\displaystyle \alpha$ and $\displaystyle \beta$, you can calculate the roots of the second equation using these.

    You can also calculate the new roots from the original $\displaystyle \alpha$ and $\displaystyle \beta$ by using...

    Sum of new roots $\displaystyle =\frac{b-2c}{a+c-b}$

    Product of new roots $\displaystyle =\frac{c}{a+c-b}$

    Given that $\displaystyle -(\alpha+\beta)=\frac{b}{a}$ and $\displaystyle \alpha\beta=\frac{c}{a}$

    you can write $\displaystyle a$ and $\displaystyle c$ in terms of $\displaystyle b$

    then solve for the sum and product of the new roots using $\displaystyle b$.

    If you do this, you will find the new roots in terms of the old ones..

    the calculations will give

    new roots=$\displaystyle \frac{\alpha}{1+\alpha},\ \frac{\beta}{1+\beta}$
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  6. #6
    Junior Member
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    Thank you Henryt999, HallsofIvy, e^(i*pi) & Archie Meade!

    I solved the equation using the methods provided by you using
    $\displaystyle
    a' = (a-b+c)
    $
    $\displaystyle
    b' = (b-2c)
    $
    $\displaystyle
    c' = c
    $

    And then used the quadratic equation formula to find the new roots in terms of $\displaystyle {a}, {\alpha}, {\beta}
    $

    The $\displaystyle a$ gets canceled, so we're left with only $\displaystyle {\alpha}, {\beta}$.

    The final answer I got is:
    $\displaystyle
    \frac{\alpha}{1+\alpha},\ \frac{\beta}{1+\beta}
    $

    Cheers.
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