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

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Karl-Olav Nyberg

Posts: 339
Registered: 12/6/04
Re: Help Needed in math
Posted: Apr 29, 2014 5:55 PM
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On Tuesday, April 29, 2014 11:46:07 PM UTC+2, konyberg wrote:
> On Tuesday, April 29, 2014 2:03:05 PM UTC+2, Peter Percival wrote:
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> > William Elliot wrote:
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> >
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> > > On Mon, 28 Apr 2014, Peter Percival wrote:
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> >
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> > >> punisher wrote:
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> >
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> > >>
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> > >>> Find the largest values of the integer a for which ax^2-5x-3 is
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> >
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> > >>> negative for all values of x.
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> > >
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> > >> Minus three.
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> > >
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> > > ax^2 - 5x - 3
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> > > (5 +- sqr(25 + 12a)/2a
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> > > 25 + 12a < 0; a < -25/12
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> > At this point one concludes that -3 is the answer.
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> > > ax^2 - 5x - 3 < -25x^2 / 12 - 5x - 3 < 0 ?
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> >
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> > > -25x^2 - 60x - 36 = -(5x + 6)^2 <= 0
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> > >
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> > --
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> >
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> > ...if someone seduced my daughter it would be damaging and horrifying
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> >
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> > but not fatal. She would recover, marry and have lots of children...
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> >
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> > On the other hand, if some elderly, or not so elderly, schoolmaster
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> > seduced one of my sons and taught him to be a homosexual, he would ruin
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> > him for life. That is the fundamental distinction. -- Lord Longford
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> Hi.
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> a < (5x + 3)/x^2
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> Find the minimum of this equation. Then find the nearest integer of this minimum from these rules.
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> a) If min(a) < 0 then a = int(a)-1
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> b) If min(a) > 0 then a = int(a)+1
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> Here min(a) = -25/12, which gives a = -3
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> KON


Sorry a = int(a)-1 if not a = int(a). Has nothing to do whatsoever if a < 0 or a > 0.
KON



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