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

**0**

# 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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