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Jan 16th, 2007, 07:17 PM
#1
Thread Starter
Member
quadratic equations
given that e, f are the roots of the equations x^2 + bx + 1 =0 and r, s are the roots of the equation x^2 + cx + 1 = 0.
Express (e - r)(e + s)(f - r)(f + s) in terms of b and c.
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Jan 17th, 2007, 01:04 PM
#2
Re: quadratic equations
Do you know how to solve for the roots of a quadratic equation?
Solve for e,f in terms of b
Solve for r,s in terms of c
Substitute your results into (e - r)(e + s)(f - r)(f + s)
And multiply it out.
If you don't know how to solve a quadratic, then:
x = [-b +/- Sqrt(b2 -4*a*c)] / 2*a
And in your case:
[e,f] = [-1 +/- Sqrt(b2-4)]/2
and
[r,s] = [-1 +/- Sqrt(c2-4)]/2
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Jan 18th, 2007, 11:32 AM
#3
Re: quadratic equations
 Originally Posted by NotLKH
Do you know how to solve for the roots of a quadratic equation?
Solve for e,f in terms of b
Solve for r,s in terms of c
Substitute your results into (e - r)(e + s)(f - r)(f + s)
And multiply it out.
If you don't know how to solve a quadratic, then:
x = [-b +/- Sqrt(b2 -4*a*c)] / 2*a
And in your case:
[e,f] = [-1 +/- Sqrt(b2-4)]/2
and
[r,s] = [-1 +/- Sqrt(c2-4)]/2
Well, this is the obvious way to go but is the "brute force" approach. I have the feling there may be some hidden trick that allows you to solve this more easily, but I may be wrong.
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Jan 18th, 2007, 06:31 PM
#4
Thread Starter
Member
Re: quadratic equations
It does get pretty hard multiplying all those numbers using that way.
{[-b + Sqrt(b2-4)]/2 - [-c + Sqrt(c2-4)]/2}{[-b + Sqrt(b2-4)]/2 +
[-c - Sqrt(c2-4)]/2} ... and so on.
And shouldnt it be [-b +/- Sqrt(b2-4)]/2 instead of [-1 +/- Sqrt(b2-4)]/2
and [-c +/- Sqrt(c2-4)]/2 instead of [-1 +/- Sqrt(c2-4)]/2 ?
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Jan 19th, 2007, 08:40 AM
#5
Re: quadratic equations
You are correct:
[e,f] = [-b +/- Sqrt(b2-4)]/2
and
[r,s] = [-c +/- Sqrt(c2-4)]/2
Alternatively, you might find these identies useful:
e+s=b
ef = 1
r+s=c
rs = 1
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Jan 23rd, 2007, 08:18 PM
#6
Thread Starter
Member
Re: quadratic equations
I think this is more correct... e+f= -b ef=1
r+s= -c rs=1
And i solved it already.
(e-r)(f-r) = ef - er - fr + r^2
= ef - r(e+f) + r^2
=1+br+r^2
Similarly (e+s)(f+s) = 1 - bs +s^2
(e+s)(f+s)(e-r)(f-r) = (1 - bs +s^2)(ef - er - fr + r^2)
Expand it and then use the identity rs= 1 to simplify it.
Then use the identity r+s=c to simplify it further.
The answer is c^2 - b^2.
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