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

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

    Once you know that, you should be able to use that to find roots of the second equation in terms of a, \alpha, and \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.
    (x-\alpha)(x - \beta) = (a-b+c)x^2 + (b-2c)x + c

    This means that

    a-b+c = 1

    \alpha \beta = c

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

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

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

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

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

    you can write a and c in terms of b

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

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

    the calculations will give

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

    I solved the equation using the methods provided by you using
    <br />
a' = (a-b+c)<br />
    <br />
b' = (b-2c)<br />
    <br />
c' = c<br />

    And then used the quadratic equation formula to find the new roots in terms of {a}, {\alpha}, {\beta}<br />

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

    The final answer I got is:
    <br />
\frac{\alpha}{1+\alpha},\ \frac{\beta}{1+\beta}<br />

    Cheers.
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