Customer Discussions > Robert Smedley forum

Question in "Introducing Pure Mathematics"

Sort: Oldest first | Newest first
Showing 1-5 of 5 posts in this discussion
Initial post: 16 Apr 2011 19:01:34 BDT
on page 426 of introducing pure mathematics, qu 10 has 2 parts. Am I supposed to answer part b with reference to part a? I can do it the long winded way but that word DEDUCE in part b has me thinking. Help

Posted on 13 Feb 2012 09:29:21 GMT
Could anyone help me with Exercise 3F, q.16b, p.113? I am a retired professor of history, aged 71, who was not allowed to do maths A level at school and is now trying to make good the gap. RJC

In reply to an earlier post on 13 Feb 2012 12:45:58 GMT
Hi I,m on holiday at the moment but will be back on saturday. I had a go at nearly all the questions in the book and have kept the answers. I'll get back to you ASAP. I did my a levels in 1969 and visited that book about 5 yrs ago. It's great isn't it.

In reply to an earlier post on 13 Feb 2012 13:07:25 GMT
Manythanks; no hurry, enjoy you holiday.

In reply to an earlier post on 17 Feb 2012 09:25:20 GMT
Last edited by the author on 17 Feb 2012 09:26:30 GMT
OK I'm home now and have the solution. You have proved part a I believe. You will have found therefore that (ax + b)/(cx +d) = (b-xd)/(cx-a). At this point, cross multiply and gather like terms and faactorise. You end up with x^2c(a+d) - x(a^2 - d^2) - b(a + d) = 0. (x^2 means x squared). Now use the quadratic formula to find the roots of that equation. It will rely on you spotting the use of the difference of 2 squares i.e. (a^2 - d^2) = (a +d) (a-d). The term under the sqare root sign can evenually be simplufied to (a+d)^2[(a-d)^2 + 4bc]. From the question, this evaluates to zero hence there is only one root therefore only one point of intersection.

I hope that this has helped. If you are still stuck I will e-mail you a copy of the solution.

Thank you for posting your comment, it has made me smile.

‹ Previous 1 Next ›
[Add comment]
Add your own message to the discussion
To insert a product link use the format: [[ASIN:ASIN product-title]] (What's this?)
Prompts for sign-in


This discussion

Discussion in:  Robert Smedley forum
Participants:  2
Total posts:  5
Initial post:  16 Apr 2011
Latest post:  17 Feb 2012

New! Receive e-mail when new posts are made.
Tracked by 2 customers