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    suppose u got
    x^2 + bx + c = 0, and the roots of the are (1 + sqrt 5) and (1 - sqrt 5)
    yeah so u have to find out the value of b and c!
    how would u go about answering this? like wat do u have to do, an explanation will do just fine.
    thanx
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    well, I'm not sure if this is right but, if 1 +- sqrt 5 has not been simplified, then

    b^2 - 4ac = 5

    If it was (2 +-sqrt 20)/2, (2 rt 5 = rt 20)

    b^2 - 4ac = 20, and b = -2.

    Maybe something along those lines?

    Sorry thats all I can think of...
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    Substitute the the roots in as values of x, you should come out two equations in terms of b and c solve them simultaneously.
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    Duh, can't believe I didn't think of that.
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    There's various ways to do it. The easiest is probably just to substitute for x and set up simultaneous equations, since you know what it is.
    Or you could use the formula equating the roots i.e

    1+ sqrt5 = (b^2-4c)/2 and
    1- sqrt5 = (b^2-4c)/2 then solve
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    (Original post by Matt_2K)
    Duh, can't believe I didn't think of that.
    Don't blame you had me thinking as well , bit of an odd question.
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    (Original post by bigbadb1319)
    suppose u got
    x^2 + bx + c = 0, and the roots of the are (1 + sqrt 5) and (1 - sqrt 5)
    yeah so u have to find out the value of b and c!
    how would u go about answering this? like wat do u have to do, an explanation will do just fine.
    thanx
    a general result about sum of roots and product of roots of quadratics solves this instantly.
    You wont know this but you can get there in the following way:
    let m=1+rt(5)
    n=1-rt(5)
    the x^2+bx+c=(x-m)(x-n)
    =x^2+(-m-n)x+mn.
    comparing coeff of x
    b=-(m+n) =>b=-2

    comparing constant term
    c=mn =>c=-4


    The general result is if ax^2+bx+c=0 has roots m,n then
    m+n=-b/a
    mn=c/a
    See if you can prove this.
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    k kl, thanx ppl


    {{{{The general result is if ax^2+bx+c=0 has roots m,n then
    m+n=-b/a
    mn=c/a
    See if you can prove this.}}}}

    i'll try that out 2...
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    Just out of interest, where are you getting these questions from?
 
 
 
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Updated: April 9, 2006
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