Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Basic algebra

  • 07-06-2012 2:21pm
    #1
    shaunie007
    Registered Users, Registered Users 2 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..


Comments

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


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


  • #3
    Michael Collins
    Moderators, Science, Health & Environment Moderators Posts: 1,852 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.


  • #4
    bnt
    Registered Users, Registered Users 2 Posts: 13,077 ✭✭✭✭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.

    You are the type of what the age is searching for, and what it is afraid it has found. I am so glad that you have never done anything, never carved a statue, or painted a picture, or produced anything outside of yourself! Life has been your art. You have set yourself to music. Your days are your sonnets.

    ―Oscar Wilde predicting Social Media, in The Picture of Dorian Gray



  • #5
    MathsManiac
    Registered Users, Registered Users 2 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.


  • #6
    mikeystipey
    Registered Users, Registered Users 2 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?


  • Advertisement
  • #7
    shaunie007
    Registered Users, Registered Users 2 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...


  • #8
    shaunie007
    Registered Users, Registered Users 2 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...


Advertisement