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Math Help - Reciprocal Roots

  1. #1
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    Reciprocal Roots

    Hey guys, having trouble with this problem :-/
    Show that the equation 2x^2-5x + 2 = 0 has roots that are reciprocals of each other. Under what conditions will a quadratic equation in the form ax^2 + bx + c = 0 have roots that are reciprocals of each other?
    Last edited by CaptainBlack; February 24th 2008 at 10:47 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Bajan View Post
    Hey guys, having trouble with this problem :-/
    you're missing a sign.

    do you mean 2x^2 + 5x + 2 = 0 or 2x^2 - 5x + 2 = 0?

    for the second part, you can expand the general form of: a(x - a) \left(x - \frac 1a \right) = 0, and see how a, b,\mbox{ and }c relate to each other. that's one way. you could also use the quadratic formula to find the roots, and then equate one to the reciprocal of the other and try to solve. things like that.

    or perhaps you can use the nature of roots to solve it. if r_1 and r_2 are the roots of the equation ax^2 + bx + c = 0, then r_1 + r_2 = - \frac ba and r1 \cdot r_2 = \frac ca.

    then plug in r_2 = \frac 1{r_1} and see what happens.

    just play around with it. tell us your findings
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Bajan View Post
    Hey guys, having trouble with this problem :-/
    Well for the quadratic 2x^2-5x+2=0 apply the quadratic formula to find the roots.


    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    you're missing a sign.

    The sign was not missing, just bad LaTeX, fixed now.

    But it did not matter the same result occurs with either sign.

    RonL
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  5. #5
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    Hello,Bajan!

    Under what conditions will a quadratic equation in the form  ax^2 + bx + c \:=\:0
    have roots that are reciprocals of each other?

    An equation with "symmetric" coefficients will have reciprocal roots . . .
    . . that is: . ax^2 + bx + a \:=\:0

    The two roots are: . x \;=\;\frac{-b + \sqrt{b^2-4a^2}}{2a}\,\text{ and }\,\frac{-b-\sqrt{b^2-4a^2}}{2a}

    . . which can be shown to be reciprocals.

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