
Re: Help Needed in math
Posted:
Apr 29, 2014 5:46 PM


On Tuesday, April 29, 2014 2:03:05 PM UTC+2, Peter Percival wrote: > William Elliot wrote: > > > On Mon, 28 Apr 2014, Peter Percival wrote: > > >> punisher wrote: > > >> > > >>> Find the largest values of the integer a for which ax^25x3 is > > >>> negative for all values of x. > > > > > >> Minus three. > > > > > > ax^2  5x  3 > > > (5 + sqr(25 + 12a)/2a > > > 25 + 12a < 0; a < 25/12 > > > > At this point one concludes that 3 is the answer. > > > > > ax^2  5x  3 < 25x^2 / 12  5x  3 < 0 ? > > > 25x^2  60x  36 = (5x + 6)^2 <= 0 > > > > > > > > >  > > ...if someone seduced my daughter it would be damaging and horrifying > > but not fatal. She would recover, marry and have lots of children... > > On the other hand, if some elderly, or not so elderly, schoolmaster > > seduced one of my sons and taught him to be a homosexual, he would ruin > > him for life. That is the fundamental distinction.  Lord Longford
Hi.
a < (5x + 3)/x^2
Find the minimum of this equation. Then find the nearest integer of this minimum from these rules. a) If min(a) < 0 then a = int(a)1 b) If min(a) > 0 then a = int(a)+1
Here min(a) = 25/12, which gives a = 3
KON

