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Thread: quadratic equations

  1. #1

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    quadratic equations

    given that e, f are the roots of the equations x^2 + bx + 1 =0 and r, s are the roots of the equation x^2 + cx + 1 = 0.

    Express (e - r)(e + s)(f - r)(f + s) in terms of b and c.

  2. #2
    pathfinder NotLKH's Avatar
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    Re: quadratic equations

    Do you know how to solve for the roots of a quadratic equation?

    Solve for e,f in terms of b
    Solve for r,s in terms of c

    Substitute your results into (e - r)(e + s)(f - r)(f + s)

    And multiply it out.

    If you don't know how to solve a quadratic, then:

    x = [-b +/- Sqrt(b2 -4*a*c)] / 2*a

    And in your case:

    [e,f] = [-1 +/- Sqrt(b2-4)]/2
    and
    [r,s] = [-1 +/- Sqrt(c2-4)]/2

  3. #3
    vbuggy krtxmrtz's Avatar
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    Re: quadratic equations

    Quote Originally Posted by NotLKH
    Do you know how to solve for the roots of a quadratic equation?

    Solve for e,f in terms of b
    Solve for r,s in terms of c

    Substitute your results into (e - r)(e + s)(f - r)(f + s)

    And multiply it out.

    If you don't know how to solve a quadratic, then:

    x = [-b +/- Sqrt(b2 -4*a*c)] / 2*a

    And in your case:

    [e,f] = [-1 +/- Sqrt(b2-4)]/2
    and
    [r,s] = [-1 +/- Sqrt(c2-4)]/2
    Well, this is the obvious way to go but is the "brute force" approach. I have the feling there may be some hidden trick that allows you to solve this more easily, but I may be wrong.
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  4. #4

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    Re: quadratic equations

    It does get pretty hard multiplying all those numbers using that way.

    {[-b + Sqrt(b2-4)]/2 - [-c + Sqrt(c2-4)]/2}{[-b + Sqrt(b2-4)]/2 +
    [-c - Sqrt(c2-4)]/2} ... and so on.

    And shouldnt it be [-b +/- Sqrt(b2-4)]/2 instead of [-1 +/- Sqrt(b2-4)]/2
    and [-c +/- Sqrt(c2-4)]/2 instead of [-1 +/- Sqrt(c2-4)]/2 ?

  5. #5
    pathfinder NotLKH's Avatar
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    Re: quadratic equations

    You are correct:

    [e,f] = [-b +/- Sqrt(b2-4)]/2
    and
    [r,s] = [-c +/- Sqrt(c2-4)]/2


    Alternatively, you might find these identies useful:
    e+s=b
    ef = 1

    r+s=c
    rs = 1

  6. #6

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    Re: quadratic equations

    I think this is more correct... e+f= -b ef=1
    r+s= -c rs=1

    And i solved it already.

    (e-r)(f-r) = ef - er - fr + r^2
    = ef - r(e+f) + r^2
    =1+br+r^2
    Similarly (e+s)(f+s) = 1 - bs +s^2

    (e+s)(f+s)(e-r)(f-r) = (1 - bs +s^2)(ef - er - fr + r^2)
    Expand it and then use the identity rs= 1 to simplify it.
    Then use the identity r+s=c to simplify it further.
    The answer is c^2 - b^2.

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