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Ramkumar Menon
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 Posted: Wed Jul 09, 2008 5:56 am Post subject: Inequalities - quadratic |
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Gurus,
I dont seem to be good at Math anymore - unfortunately I require a
solution critical to a problem I am facing.
What is the range of x for
x^2 - p < x
where p > 0 and x > 0.
Please help!
Ram |
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Torsten Hennig
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| Posted: Wed Jul 09, 2008 7:18 am Post subject: Re: Inequalities - quadratic |
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| Quote: |
Gurus,
I dont seem to be good at Math anymore - unfortunately I >require a
solution critical to a problem I am facing.
What is the range of x for
x^2 - p < x
where p > 0 and x > 0.
Please help!
Ram
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First determine the zeros of the quadratic polynomial
f(x) = x^2 - x - p.
Then, since the parabola is open above, the open
interval between the zeros satisfies your above
inequality.
(I get x1=0.5*(1-sqrt(1+4*p)) and x2=0.5*(1+sqrt(1+4*p))
as zeros of the quadratic polynomial and so
x-values within the open interval (x1;x2) satisfy
your inequality).
Best wishes
Torsten. |
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Adrian Duma
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| Posted: Wed Jul 09, 2008 7:54 am Post subject: Re: Inequalities - quadratic |
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| Quote: |
Gurus,
I dont seem to be good at Math anymore -
unfortunately I require a
solution critical to a problem I am facing.
What is the range of x for
x^2 - p < x
where p > 0 and x > 0.
Please help!
Ram
|
Your inequality is equivalent to:
x^2 - x + 1/4 < p + 1/4 (x > 0).
Remark that x^2 - x + 1/4 = (x - 1/2)^2, so that we obtain
(x - 1/2)^2 < p + 1/4 (x > 0).
This is, in turn, equivalent to:
|x - 1/2| < sqrt(p + 1/4) (x > 0).
Therefore, the range of x (for a fixed p > 0) shall be the intersection:
(1/2 - sqrt(p + 1/4), 1/2 + sqrt(p + 1/4)) /\ (0, +oo),
that is precisely the open interval
(0, 1/2 + sqrt(p + 1/4)). |
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William Elliot
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| Posted: Wed Jul 09, 2008 11:04 am Post subject: Re: Inequalities - quadratic |
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On Tue, 8 Jul 2008, Ramkumar Menon wrote:
| Quote: |
What is the range of x for
x^2 - p < x
where p > 0 and x > 0.
x^2 - x - p = 0 |
y = x - 1; -1 < y
y^2 + y - p = 0
y = (-1 +- sqr(1 + 4p))/2 = (-1 + sqr(1 + 4p))/2 = c
c^2 + c = p
If c <= y, then y^2 + y >= p.
If 0 <= y < c, then y^2 + y < p.
If -1 < y < 0, then y^2 + y < 1
Thus if p < 1, discard y in (-1,0).
and if 1 <= p, included y in (-1,0)
Convert back to x. |
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José Carlos Santos
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| Posted: Wed Jul 09, 2008 11:04 am Post subject: Re: Inequalities - quadratic |
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On 09-07-2008 6:56, Ramkumar Menon wrote:
| Quote: |
I dont seem to be good at Math anymore - unfortunately I require a
solution critical to a problem I am facing.
What is the range of x for
x^2 - p < x
where p > 0 and x > 0.
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0 < x < 1/2 + sqrt(1/4 + p)
Best regards,
Jose Carlos Santos |
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Ramkumar Menon
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| Posted: Thu Jul 10, 2008 2:00 am Post subject: Re: Inequalities - quadratic |
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On Jul 9, 7:09 am, Ray Vickson <RGVick...@shaw.ca> wrote:
| Quote: |
On Jul 8, 10:56 pm, Ramkumar Menon <ramkumar.me...@gmail.com> wrote:
Gurus,
I dont seem to be good at Math anymore - unfortunately I require a
solution critical to a problem I am facing.
What is the range of x for
x^2 - p < x
where p > 0 and x > 0.
Please help!
Ram
You want f(x) < 0, where f(x) = x^2 - x - p. Solve the quadratic
equation f(x) = 0; you will get two roots, r1 and r2, with r1 < 0 and
r2 > 0. (You can find r1 and r2 using the familiar quadratic formula.)
The function f(x) is < 0 at x = 0 (do you see why?), so the interval
r1 < x < r2 is the region where f < 0, which is what you want;
actually, you want the part having only x > 0.
R.G. Vickson
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Thanks everyone for the prompt responses.
Ram |
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