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# c1 question watch

1. 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
2. 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...
3. Substitute the the roots in as values of x, you should come out two equations in terms of b and c solve them simultaneously.
4. Duh, can't believe I didn't think of that.
5. 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
6. (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.
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.
8. 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...
9. Just out of interest, where are you getting these questions from?

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Updated: April 9, 2006
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