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Math Help - polynomials

  1. #1
    may
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    Exclamation polynomials

    if the root of the equation 3x^2-6x+1=0 are a and b,find the value of

    [a^2+1/(b^2)][b^2+1/(a^2)]
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  2. #2
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    Quote Originally Posted by may View Post
    if the root of the equation 3x^2-6x+1=0 are a and b,find the value of

    [a^2+1/(b^2)][b^2+1/(a^2)]
    First of all, I can't tell if this is

    \left(\frac{a^2 + 1}{b^2}\right)\left(\frac{b^2 + 1}{a^2}\right) or \left(a^2 + \frac{1}{b^2}\right)\left(b^2 + \frac{1}{a^2}\right).


    Anyway, to begin to answer your questions...

    3x^2 - 6x + 1 = 0

    x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(1)}}{2(3)}

     = \frac{6 \pm \sqrt{36 - 12}}{6}

     = \frac{6 \pm \sqrt{24}}{6}

     = \frac{6 \pm 2\sqrt{6}}{6}

     = \frac{3 \pm \sqrt{6}}{3}.


    So a = \frac{3 - \sqrt{6}}{3} and b = \frac{3 + \sqrt{6}}{3}. Substitute these values into your expression and simplify.
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