quadratic equations

[fiery123]

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.

[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

[krtxmrtz]

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.

[fiery123]

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 ?

[NotLKH]

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

[fiery123]

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.