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Topic: Help Needed in math
Replies: 25   Last Post: May 2, 2014 6:01 PM

 Messages: [ Previous | Next ]
 Karl-Olav Nyberg Posts: 1,575 Registered: 12/6/04
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^2-5x-3 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

Date Subject Author
4/28/14 punisher
4/28/14 quasi
4/29/14 Peter Percival
4/29/14 Karl-Olav Nyberg
4/28/14 Peter Percival
4/29/14 William Elliot
4/29/14 Peter Percival
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 William Elliot
4/30/14 Karl-Olav Nyberg
4/30/14 William Elliot
4/30/14 Karl-Olav Nyberg
4/30/14 William Elliot
5/1/14 Karl-Olav Nyberg
5/1/14 William Elliot
5/1/14 quasi
5/2/14 William Elliot
5/2/14 quasi
5/2/14 William Elliot
5/2/14 snmpprotocol@gmail.com
5/2/14 quasi
5/2/14 Karl-Olav Nyberg
5/2/14 quasi