The Student Room Group

Solve this?

Does anybody know how to solve for rr?

5000+50001+r+500(1+r)2+500(1+r)3=5000+5001+r+500(1+r)2+6000(1+r)3-5000+\frac{5000}{1+r}+\frac{500}{(1+r)^2}+\frac{500}{(1+r)^3}=-5000+\frac{500}{1+r}+\frac{500}{(1+r)^2}+\frac{6000}{(1+r)^3}

Thanks!
Reply 1
It's a cubic in (r+1). I can't see an obvious way to solve it, so multiply through, collect and factorise?

(Actually, cancelling the -5000 makes it a quadratic in (r+1), so use the formula if it doesn't factorise)
Reply 2
Original post by SimonM
It's a cubic in (r+1). I can't see an obvious way to solve it, so multiply through, collect and factorise?

(Actually, cancelling the -5000 makes it a quadratic in (r+1), so use the formula if it doesn't factorise)


Nope, I can't do it. I'm an economist (read: second-rate mathematician). Has anybody been able to do this?
Let u = (r+1) then multiply through by u^3. You'll get a quadratic in u (or r+1).
Reply 4
Original post by dave_chapman
Let u = (r+1) then multiply through by u^3. You'll get a quadratic in u (or r+1).


I don't know what that means. Can you show me?
Reply 5
50001+r+500(1+r)2+500(1+r)3=5001+r+500(1+r)2+6000(1+r)3/(1+r)3\frac{5000}{1+r}+\frac{500}{(1+r)^2}+\frac{500}{(1+r)^3}=\frac{500}{1+r}+\frac{500}{(1+r)^2}+ \frac{6000}{(1+r)^3} / (1+r)^3 - multiply by (1+r)3(1+r)^3 provided that r is not equal -1.
5000(1+r)31+r+500(1+r)3(1+r)2+500(1+r)3(1+r)3=500(1+r)31+r+500(1+r)3(1+r)2+6000(1+r)3(1+r)3\frac{5000(1+r)^3}{1+r}+\frac{500(1+r)^3} {(1+r)^2}+\frac{500(1+r)^3}{(1+r)^3}=\frac{500(1+r)^3}{1+r}+ \frac{500(1+r)^3}{ (1+r)^2}+\frac{6000(1+r)^3}{(1+r)^3}
5000(1+r)2+500(1+r)+500=500(1+r)2+500(1+r)+60005000(1+r)^2+500(1+r)+500=500(1+r)^2+500(1+r)+6000 - now cancel out the identical bits on both sides:
5000(1+r)2+500=500(1+r)2+60005000(1+r)^2+500=500(1+r)^2+6000 - now get everything on one side of the equation
4500(1+r)25500=04500(1+r)^2-5500=0 - now you can substitute (1+r) with u and solve this quadratic equation for u (u is not equal to 0)
Is this some sort of DCF?
(edited 13 years ago)

Quick Reply

Latest