maths problem - please help!!

Multiply the rhs out and put it in the form of a standard quadratic eqn, ax^2 + Bx +c =0 and use the standard solution.

Ed gave the outline:

so in this case.... c^2 = 4h(2R - h)

becomes

The quadratic formula solves for an equation in this form: ax^2 + bx + c = 0

For which, the solutions are x = [ -b +- sqrt(b^2 - 4ac) ] / 2a

Therefore, you rename the terms in a b and c

In this case, a is 4 b is -8R and c is c^2

You insert these substitutions in the x = equation to find the two solutions... -b PLUS something all over something and -b MINUS something all over something

OK?

Brian W (There are better ways to do homework than ask on a newgroup...)

thanks. homework????? you are kidding aren't you? it's been 20 years since l've done homework, hence the question.

hmmm...it's been 50 years since I did homework, but I found another way to answer your question. This is something you could do too. It's quicker my way!

Brian W

i got very good and prompt replies from this message board. if i'd been in my office then agreed, i could have easily asked a colleague (although i didn't really want to expose my lack of basic maths!!) or checked my engineering/maths handbooks. as it was i was on the road, with a 101 things going on, i just needed to get an answer without having to spend too much effort finding it out for myself and without the usual resources to hand. many people post messages where they could probably get the same information by searching through books, using internet resources or asking colleagues. message boards are just another method.

anyway, thanks for the answer. it helped me out just when i needed it.