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Roots of a quadratic equation

  1. Sep 30, 2013 #1
    Solve the following quadratic equation. Use factorisation if possible.

    X2 - 4X - 8 = 0


    Normally I wouldn't have trouble factorising a quadratic, but I have just been introduced to a new way to do it and I want to use this way to answer the question.

    Here's how far I get, then i'm unsure what to do with the info I have.



    X2 - 4X - 8 = 0

    Let [itex]\alpha[/itex] and [itex]\beta[/itex] be the roots of the equation.

    (X-[itex]\alpha[/itex])(X-[itex]\beta[/itex]) = 0

    Therefore

    X2 - ([itex]\alpha[/itex]+[itex]\beta[/itex])X + [itex]\alpha[/itex][itex]\beta[/itex] = 0

    Or in another form;

    X2 + [itex]\frac{b}{a}[/itex]X + [itex]\frac{c}{a}[/itex] = 0

    Comparing coefficients.

    [itex]\alpha[/itex]+[itex]\beta[/itex] = -[itex]\frac{b}{a}[/itex] = 4

    [itex]\alpha[/itex][itex]\beta[/itex] = [itex]\frac{c}{a}[/itex] = -8

    And now i'm confused about what I can do with this info to find the factors of the original quadratic.

    Any help is appreciated!

    Thanks.
     
  2. jcsd
  3. Sep 30, 2013 #2

    tiny-tim

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    Hi BOAS! :smile:
    So ##\alpha## = 2 + C, ##\beta## = 2 - C. :wink:

    (or you could simply complete the square)
     
  4. Sep 30, 2013 #3
    I don't follow how you got there, please could you elaborate?

    I tried completing the square to see if it would make anything clear to me, but I get x = ±√12 - 2 which doesn't seem right :confused:
     
  5. Sep 30, 2013 #4

    tiny-tim

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    Hi BOAS! :smile:
    If you define C by ##\alpha## = 2 + C, then ##\beta## = … ? :wink:
    you should have got + 2 :redface:
     
  6. Sep 30, 2013 #5
    Where does that definition come from?

    Oops - careless mistake.
     
  7. Sep 30, 2013 #6

    tiny-tim

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    Isn't it obvious?

    If A + B = N, then A and B must be equidistant from N/2 ?
     
  8. Sep 30, 2013 #7
    <delete>
     
    Last edited: Sep 30, 2013
  9. Sep 30, 2013 #8

    epenguin

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    If you have been given a new method for solving use that. But it can't be what you started doing, i.e. to use these relations between roots and coefficients. Which maybe are what is new to you.

    The reason is these relations are symmetric. E.g. in your equations there is nothing to distinguish α from β. (Your expressions are not changed by interchanging α and β.) So you might find a method to eliminate β and end up with an equation with α only. But because youve started with completely symmetrical expressions the same method has got to give you an equation for β - in fact it will be the same equation. You have only changed the name you give to 'a root'. α or β or x. You will just get your starting equation. Just in case you've tried and notice it is like trying to pin down a globule of mercury, this is why. :smile:

    In the case you have extra information, if e.g you are told one root is twice the other or other relation between roots, then you can use these relations in the solution - or rather the reduction of degree. But not in the general case.

    Added: Posts crossed - you did what I mentioned but if you check out a mistake you made in calculation in the last post you will get an equation in β that is only the starting equation! :tongue:
     
    Last edited: Sep 30, 2013
  10. Sep 30, 2013 #9
    You're correct, I haven't been given a new method for solving equations - The relationship between coefficients is what's new to me.

    It sounds like i'd be much better off simply completing the square and using this method for questions that actually require it, such as "find the equation whose roots are the inverse of ..."
     
  11. Sep 30, 2013 #10

    epenguin

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    Right. But it was an idea, and most people who are thinking actively and meet these relations try a bit to go down that road, which unfortunately leads back to the starting point! :biggrin:
     
  12. Sep 30, 2013 #11
    Trial and error :)
     
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