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Math Help - Finding equation using roots

  1. #1
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    Finding equation using roots

    Hi, I'm having trouble with this question: Find the equation whose roots are the squares of the reciprocals of the roots of x^2+6x-2.

    So far, I've done:

    Let x1 and x2 represent roots of the original equation.

    x1 + x2= -b/a
    = -6
    (x1)(x2) = c/a
    = -2

    Then I figured the roots of the required equation would be (1/x1^2) and (1/x2^2)

    So the product of the required equation would be 1/4. However, I'm having difficulty finding the sum. I think it should be:

    x1^2 + x2^2
    ___________

    4

    However, I don't know how to fix the numerator so that I can sub a value into it and get the correct sum. (Then obviously, I can just sub the sum and product into x^2-x[sum]+product). I thought of square-rooting the entire equation, but that doesn't work out. Any help finding the sum for the required equation would be greatly appreciated Thank you!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by starswept View Post
    Hi, I'm having trouble with this question: Find the equation whose roots are the squares of the reciprocals of the roots of x^2+6x-2.

    So far, I've done:

    Let x1 and x2 represent roots of the original equation.

    x1 + x2= -b/a
    = -6
    (x1)(x2) = c/a
    = -2

    Then I figured the roots of the required equation would be (1/x1^2) and (1/x2^2)

    So the product of the required equation would be 1/4. However, I'm having difficulty finding the sum. I think it should be:

    x1^2 + x2^2
    ___________

    4

    However, I don't know how to fix the numerator so that I can sub a value into it and get the correct sum. (Then obviously, I can just sub the sum and product into x^2-x[sum]+product). I thought of square-rooting the entire equation, but that doesn't work out. Any help finding the sum for the required equation would be greatly appreciated Thank you!
    i don't remember the rules for the sum and products of roots and all that stuff, but i think i have another method that can work, if you're interested

    or if you have to use that method, i'll look up those rules.

    this site seems to be sufficient
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  3. #3
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    Say that x^2  + 6x - 2 = 0 has roots p & q.
    Then p + q =  - 6,\quad pq =  - 2 which implies p^2  + q^2  = 36 - 2pq = 40.
    Now the equation you want is \left( {x - \frac{1}{{p^2 }}} \right)\left( {x - \frac{1}{{q^2 }}} \right) = x^2  - \left( {\frac{{p^2  + q^2 }}{{p^2 q^2 }}} \right)x + \frac{1}{{p^2 q^2 }} = 0.

    Can you finish?
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    MHF Contributor red_dog's Avatar
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    \displaystyle \frac{1}{x_1^2}+\frac{1}{x_2^2}=\frac{x_1^2+x_2^2}  {x_1^2x_2^2}=\frac{(x_1+x_2)^2-2x_1x_2}{(x_1x_2)^2}
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    Say that x^2  + 6x - 2 = 0 has roots p & q.
    Then p + q =  - 6,\quad pq =  - 2 which implies p^2  + q^2  = 36 - 2pq = 40.
    Now the equation you want is \left( {x - \frac{1}{{p^2 }}} \right)\left( {x - \frac{1}{{q^2 }}} \right) = x^2  - \left( {\frac{{p^2  + q^2 }}{{p^2 q^2 }}} \right)x + \frac{1}{{p^2 q^2 }} = 0.

    Can you finish?
    i was thinking of doing something like that. nice!
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  6. #6
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    Quote Originally Posted by red_dog View Post
    \displaystyle \frac{1}{x_1^2}+\frac{1}{x_2^2}=\frac{x_1^2+x_2^2}  {x_1^2x_2^2}=\frac{(x_1+x_2)^2-2x_1x_2}{(x_1x_2)^2}
    Just one question about this; how did you get (x1+x2)^2 - 2x1x2? That gives the correct answer, but I worked it out to be (x1+x2)^2 + 2x1x2. I'm just not certain how to get the minus 2x1x2.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by starswept View Post
    Just one question about this; how did you get (x1+x2)^2 - 2x1x2? That gives the correct answer, but I worked it out to be (x1+x2)^2 + 2x1x2. I'm just not certain how to get the minus 2x1x2.

    (x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2

    \Rightarrow x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2
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