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Math Help - Root of the equation?

  1. #1
    Super Member fardeen_gen's Avatar
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    Root of the equation?

    Find all the pairs (a,b) of real numbers such that whenever
    α is a root of x^2 + ax + b = 0, α^2 - 2 is also a root of the equation.
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  2. #2
    Member Henderson's Avatar
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    Are the a's all meant to be the same variable?

    The first step either way would be to recognize that α(α^2 - 2) = b and (α) + (α^2 - 2) = -a.
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by fardeen_gen View Post
    Find all the pairs (a,b) of real numbers such that whenever
    α is a root of x^2 + ax + b = 0, α^2 - 2 is also a root of the equation.
    Rought method :
    x^2+ax+b can be factored into (x-x_1)(x-x_2)=0 where x_1 and x_2 are the roots of the equation.

    --> x^2+ax+b=(x-a)(x-(a^2-2))=(x-a)(x-a^2+2)

    Developping the RHS, we get : x^2-ax-a^2x+2x+a^3-2a=x^2-x(a^2+a-2)+a^3-2a

    --> a^2+a-2=-a and a^3-2a=b

    --------------------------
    Other method that will end up at the same point as above

    In an equation x+mx+n=0, the sum of the roots is -m and the product of the roots is n.

    Here, we have x+ax+b=0 and roots a and a-2.

    Therefore (a+a-2)=-a and a^3-2a=b

    --------------------------

    a+a-2=-a ------> a+2a-2=0.
    Completing the square :

    a+2a+1-3=0 ------> (a+1)=3 -----> a+1=+ or - sqrt(3) ----> a_1=-1+ \sqrt{3} and a_2=-1-\sqrt{3}


    And then calculate b_1=a_1^3-2a_1 and b_2=a_2^3-2a_2
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  4. #4
    Super Member fardeen_gen's Avatar
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    So what are the pairs of numbers?
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  5. #5
    Moo
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    It's all computation...

    b_1=a_1^3-2a_1=(-1+ \sqrt{3})^3-2(-1+ \sqrt{3})=\dots=4 \sqrt{3}-8

    So the first couple is (-1+ \sqrt{3}~,~4 \sqrt{3}-8)

    Do the same with the second
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