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Basic algebra

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  • 07-06-2012 3:21pm
    #1
    shaunie007
    Registered Users Posts: 79 ✭✭


    If x = 1 - 2^1/2 is a root of the function ax^2 + bx + c = 0,where a,b,c are an element of R what is the other root?

    Now, I know that the root is x = 1 + 2^1/2, is the other root, but I can't explain it...

    Any help, much appreciated..


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    #2
    red_fox
    Registered Users Posts: 642 ✭✭✭red_fox


    It's all about how roots are related to factors. (This is meant as a hint rather than a solution)


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    #3
    Michael Collins
    Moderators, Science, Health & Environment Moderators Posts: 1,846 Mod ✭✭✭✭Michael Collins


    shaunie007 wrote: »
    ...Now, I know that the root is x = 1 + 2^1/2, is the other root...

    This isn't actually the case when your coefficients can be among the real numbers. If they were limited to the rationals then, yes, this would be correct.


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    #4
    bnt
    Registered Users Posts: 12,961 ✭✭✭✭bnt


    If you think about the formula for the roots of a quadratic equation, and the form of results it gives you, then you can get that answer by assuming a = 1/2, substituting the other values above the line, and changing sign where the formula has ±.

    However, it's not the only possible answer, since there are all kinds of combinations of (a, b, c) that would produce a quadratic curve with that one given root. I did a little playing in Excel's "Goal Seek" function and it wasn't hard to find others. If the first root is always 1-sqrt(2) = -0.414, then
    (a, b, c) = (0.5, -4.624, 2) and the second root = 9.662
    (a, b, c) = (-1.5, -2.5, -0.778) and the second root = -1.253
    and so on.

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    #5
    MathsManiac
    Registered Users Posts: 1,595 ✭✭✭MathsManiac


    bnt wrote: »
    ...
    However, it's not the only possible answer, since there are all kinds of combinations of (a, b, c) that would produce a quadratic curve with that one given root. I did a little playing in Excel's "Goal Seek" function and it wasn't hard to find others. If the first root is always 1-sqrt(2) = -0.414, then
    (a, b, c) = (0.5, -4.624, 2) and the second root = 9.662
    (a, b, c) = (-1.5, -2.5, -0.778) and the second root = -1.253
    and so on.

    Indeed. In fact, the second root can be any real number you like. Given any real number k as the second root, the following choice of a, b, and c gives a quadratic whose roots are 1-sqrt(2) and k:
    a = 1
    b = sqrt(2) - 1 - k
    c = (1 - sqrt(2))k.

    In fact, there are uncountably many different sets of values for a, b, and c that give a quadratic whose roots are 1-sqrt(2) and k.


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    #6
    mikeystipey
    Registered Users Posts: 183 ✭✭mikeystipey


    Yes afaik there are an infinite number of solutions to this problem, i.e. you could pair any real number value for x with the other value to form a quadratic equation. So interested to know where you got this question from OP?


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    #7
    shaunie007
    Registered Users Posts: 79 ✭✭shaunie007


    mikeystipey wrote: »
    Yes afaik there are an infinite number of solutions to this problem, i.e. you could pair any real number value for x with the other value to form a quadratic equation. So interested to know where you got this question from OP?

    Froma fairly unreliable mock paper.

    Cheers folks for your answers...


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    #8
    shaunie007
    Registered Users Posts: 79 ✭✭shaunie007


    mikeystipey wrote: »
    Yes afaik there are an infinite number of solutions to this problem, i.e. you could pair any real number value for x with the other value to form a quadratic equation. So interested to know where you got this question from OP?

    From a fairly unreliable mock paper.

    Cheers folks for your answers...


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