Hi all,
Just stumbled accross this site so a quick hi to all here! Anyway on with my question if I may!
Higher Maths - 2000 - Paper One - Question 5. I'd type it out but I cant find some of the characters for the recurrance relation formula!
I was thinking simultaneous equations but I'm confuzzled and the answers just dont help at all. lol.
Anyway thanks anyone for any help!
Paolo


2000 Paper One Q5 (Non Calc!)
Started by Max018, May 09 2004 07:18 PM
3 replies to this topic
#1
Posted 09 May 2004 - 07:18 PM
#2
Posted 09 May 2004 - 07:46 PM
For the first, L = b / (1-a) = 10 / (1 - a)
For the second, L = 16 / (1 - a^2)
Equate these:
16 / (1 - a^2) = 10 / (1 - a)
16 (1 - a) = 10 (1 - a^2)
16 - 16a = 10 - 10a^2
16 - 16a - 10 + 10a^2 = 0
10a^2 - 16a + 6 = 0
5a^2 - 8a + 3 = 0
Using the quadratic equation
a = [ -b +/- (b^2-4ac) ] / 2a = (8 +/- 2 ) / 10 = 1 or 3/5
(I get 2 solutions but the answer only gives 3/5. Don't see why the other isn't valid).
So substitute into the first recurrance relation to get L = 10 / (1 - 3/5) = 25
For the second, L = 16 / (1 - a^2)
Equate these:
16 / (1 - a^2) = 10 / (1 - a)
16 (1 - a) = 10 (1 - a^2)
16 - 16a = 10 - 10a^2
16 - 16a - 10 + 10a^2 = 0
10a^2 - 16a + 6 = 0
5a^2 - 8a + 3 = 0
Using the quadratic equation
a = [ -b +/- (b^2-4ac) ] / 2a = (8 +/- 2 ) / 10 = 1 or 3/5
(I get 2 solutions but the answer only gives 3/5. Don't see why the other isn't valid).
So substitute into the first recurrance relation to get L = 10 / (1 - 3/5) = 25
#3
Posted 09 May 2004 - 08:06 PM
Hi,
Ah, brilliant! Yeah I got to the cross multiply bit but got a bit stumped after that. Could you explain a wee bit why you used b^2 - 4ac to get the two answers, I dont get this line:
Ah, brilliant! Yeah I got to the cross multiply bit but got a bit stumped after that. Could you explain a wee bit why you used b^2 - 4ac to get the two answers, I dont get this line:
QUOTE |
Using the quadratic equation a = [ -b +/- (b^2-4ac) ] / 2a = (8 +/- 2 ) / 10 = 1 or 3/5 |
I'd problably get it if I wrote it down, looks so different on the PC!
*edit*
Ah, I'd have used the quadratic equation (5a-3)(a-1), thats where we went different. And the 1 isnt valid because in the limit formula -1 < a < 1, I get it now!
*/edit*
Thanks a ton,
Paolo

#4
Posted 09 May 2004 - 09:34 PM
I didn't notice that it factorised easily, so I used the quadratic formula as here:
http://www.sosmath.com/algebra/quadraticeq...draformula.html
(in my message above b^2-4ac should have been (b^2-4ac)^1/2
http://www.sosmath.com/algebra/quadraticeq...draformula.html
(in my message above b^2-4ac should have been (b^2-4ac)^1/2
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