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Math Help - find the cubic polynomial

  1. #1
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    find the cubic polynomial

    Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3.
    Find the cubic polynomial whose roots are
    a) x₁^2, x₂^2, x₃^2
    b) 1/(x₁), 1/(x₂), 1/(x₃)
    Because we are learning resolvent cubic now, so I think that this may relate to it. Can someone tell me how to find them?
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  2. #2
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    Quote Originally Posted by apple2009 View Post
    Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3.
    Find the cubic polynomial whose roots are
    a) x₁^2, x₂^2, x₃^2
    b) 1/(x₁), 1/(x₂), 1/(x₃)
    Because we are learning resolvent cubic now, so I think that this may relate to it. Can someone tell me how to find them?
    I tell you the technique

    Let  y = x_1^2 as a root of the new polynomial we are looking for . The key is to write the original root as the subject . ie.  x_1 = \sqrt{y}

    Then sub. back to the equation given :

     \sqrt{y}^3 + 7y - 8\sqrt{y} + 3  = 0

     \sqrt{y} ( y-8) = -(7y+3)

    Squaring gives :

     y(y-8)^2 = (7y+3)^2 I leave the expansion to you .

    Try to do the second one by this method .
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  3. #3
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    Quote Originally Posted by apple2009 View Post
    Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3. Find the cubic polynomial whose roots are x₁^2, x₂^2, x₃^2
    Using Vieta's relations, we see that x_{1}+x_{2}+x_{3} = -7, img.top {vertical-align:15%;} x_{1}x_{2}x_{3} = -3. So [Math] x_{1}^{2}+x_{2}^{2}+x_{3}^2= (x_{1}+x_{2}+x_{3})^2-2\left(x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2}\right) = (-7)^2-2(-8) = 65
    " alt="x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2} = -8[/Math], and x_{1}x_{2}x_{3} = -3. So [Math] x_{1}^{2}+x_{2}^{2}+x_{3}^2= (x_{1}+x_{2}+x_{3})^2-2\left(x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2}\right) = (-7)^2-2(-8) = 65
    " /> and x_{1}^2x_{2}^2+x_{1}^2x_{3}^2+x_{3}^2x_{2}^2 = (x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2})^2-2(x_{1}x_{2}x_{3})(x_{1}+x_{2}+x_{3}) = (-8)^2-2(-3)(-7) = 22 and x_{1}^2x_{2}^2x_{3}^2 = (x_{1}x_{2}x_{3})^2 = (-3)^2 = 9 , which gives x^3+65x^2+22x+9. The signs alternate, so the required equation is [LaTeX ERROR: Convert failed] , which simplependulum's equality yields as well.
    Last edited by TheCoffeeMachine; June 6th 2010 at 12:49 PM.
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