quadratic polynomial

niharika_kumar · 2015-04-24 23:18:27

Let P(x) be a quadratic polynomial with real coefficients such that for all real x the relation 2(1+P(x))=P(x-1)+P(x+1) holds.
P(0)=8 and P(2)=32.
If the range of P(x) is [m,∞), then the value of m is?

Bob · 2015-04-24 23:31:21

hi Niharika,

You could proceed like this:

Use the 'P' recurrence relationship to work out P(1).

Then call the quadratic ax^2 + bx + c, substitute in the known values when x = 0, 1 and 2 and solve for a, b, and c.

Once you know the quadratic you'll see it has a positive x^2 so it is 'U' shaped rather than the other way up. So it has a minimum value. The minimum for a quadratic is always at -b/(2a) and all you need is the 'y' coordinate at that minimum point.

Hope that helps,

Bob

Olinguito · 2015-04-25 00:33:22

straightaway.

niharika_kumar · 2015-05-01 15:09:12

Here is one more question on quadratic equations.

Assume that p is a real number.Find the possible values of p in order to have real solutions for

.

I shifted x^1/3 to RHS and cubed both sides and tried to simplify it to a 2 degree equation so that I could apply D≥0 for real roots, but was unsuccessful in making the equation quadratic.
Pls help.

bobbym · 2015-05-01 17:07:04

Hi;

I tried the substitution x = y^3 and that found the value of the discriminant for p but I am not sure about that idea.

Bob · 2015-05-01 20:10:06

hi Niharika

I started by having a look at a graph. By using 'a' instead of 'p' you can use the slider to try varying 'a'.

You'll see they all look similar with the intersection moving right as 'a' gets bigger.

Here's the graph when a = 2.

So then I did this:

where y = x^(1/3)

So

We need y ≥ 0 so

That seems to agree with the graphical results.

Bob

niharika_kumar · 2015-05-02 03:54:25

Well the answer given is p≥ -1/4.

bobbym · 2015-05-02 04:37:48

That is the answer I got but I do not like it.

Bob · 2015-05-02 05:09:49

That appears to come from the discriminant of the y quadratic. But you still have to find x values from those ys.

If p = -0.25 then y = -0.5 and so x = -0.7937.

In the original equation the LHS = 0.10969 which is not 1. So that is not a solution.

Bob

Olinguito · 2015-05-02 23:06:06

bob bundy wrote:

If p = -0.25 then y = -0.5 and so x = -0.7937.

Olinguito · 2015-05-02 23:18:22

niharika_kumar wrote:

.

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Last edited by Olinguito (2015-05-04 00:09:46)

Bob · 2015-05-05 00:08:47

hi

Many thanks Olinguito, for finding my error. I had done cube root of y to get x, instead of cube y to get x.

I was still puzzled about why my graph didn't show any negative x values. I experimented with the grapher for some time but couldn't get a reliable graph. As negative values will have a real cube root there should be a graph for negative x. It seems to be the way the grapher has been programmed.

So I used Excel (sorry folks ) to generate values, checked a few by long hand, and when satisfied it was working out ok I made a scatter graph of the points. That's the best graph I think, for this data.

Here's a screen shot of two graphs with p = -0.24 and -0.26 and then the critical one with p = -0.25.

Bob

Olinguito · 2015-05-05 10:34:53

You are taking values of x and p independently of each other – but they are not independent of each other! x and p are related by

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In other words your <cube root c> minus <cube root a> must be equal to 1 – but as you can see they are not.

Last edited by Olinguito (2015-05-05 14:08:21)

Olinguito · 2015-05-05 10:57:52

PS:

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so maybe this is the graph you want to plot: http://www.wolframalpha.com/input/?i=pl … a=%5E_Real

Last edited by Olinguito (2015-05-05 14:09:18)

Bob · 2015-05-05 21:14:24

hi Olinguito

I was deliberately taking x and p as independent; looking at the graph for a particular p, and finding which x (if any) makes the expression come to 1. I explored a whole family of graphs with varying p and could see that as p approached -0.25 from above the graph dropped lower. At p = -0.24 the graph just crosses the line y = 1. At p = -0.25 it just touches y = 1; and at p = -0.26 the curve fails to rise high enough.

I realise that isn't a proof, but I like to get a visual demonstration of a result. Before your proof others were unsure if p = -0.25 was the correct result. I did the graphs to show it is plausible. I'm following the bobbym's mantra to try it with numbers .

Bob

bobbym · 2015-05-06 07:18:55

bob bundy wrote:

Before your proof others were unsure if p = -0.25 was the correct result.

I am sorry for my comment without an explanation but I was unsure of the answer I got. For one thing, I do not like the way the problem is worded.

Find the possible values of p in order to have real solutions for

I am bothered by the word "possible."

Synonyms for possible: conceivable, plausible, imaginable, thinkable, believable, likely, potential, probable, credible, tenable, odds-on

Holmes wrote:

I am accustomed to have mystery at one end of my cases, but to have it at both ends is too confusing.

I am wary of problems that have any ambiguity in the phrasing. Anyway, that led me to try to find the real roots of that equation when p = - 1 / 4 or ( - 1 / 4 ) <= p < 0. I was unable to find any real solutions. Wolfram can not either unless you make a choice between principal value of the root or real value of the root. One produces an answer, the other does not.

niharika_kumar · 2015-05-07 01:15:53

well I came up with the same solution yesterday as olinguito.
It was quietly easy.
Thanks for helping me out.

Bob · 2015-05-07 03:32:11

bobbym wrote:

( - 1 / 4 ) <= p < 0. I was unable to find any real solutions.

Does that mean that my graph for p = -0.24 in post 12 is wrong?

Bob

bobbym · 2015-05-07 09:29:58

Hi;

I truly do not know, that is why I said I did not like my answer.

Bob · 2015-05-07 10:23:12

hi bobbym,

If p = -0.24 and x = -0.064

x + 3p + 1 = - 0.064 - 0.72 + 1 = 0.216 so cube root = 0.6

and cube root of x = - 0.4 so LH expression = 0.6 -- 0.4 = 1

Bob

bobbym · 2015-05-07 12:23:13

Hi Bob;

It depends on how you interpret the cube root. Wolfram believes there are two interpretations. A principal cube root and a real valued one, just like with square roots. It is my understanding that the default one is the principal cube root. Using that with your p and q will return a complex number and not 0.

Olinguito · 2015-05-07 15:06:38

When you allow complex values, then every real number will have exactly three cube roots: one of them is real and the other two conjugate complex numbers. If you want only real solutions, you should tell Wolfram explicitly – otherwise it will give you its "principal value", which is not always the real root.

Last edited by Olinguito (2015-05-07 15:10:32)

Olinguito · 2015-05-07 15:18:10

Wolfram always returns a complex number as the principal value of the cube root of a negative real number. Presumably this is the complex number in the first quadrant of the Argand diagram.

Last edited by Olinguito (2015-05-07 15:27:54)

Olinguito · 2015-05-07 20:11:31

http://www.mathisfunforum.com/viewtopic.php?id=22191

Math Is Fun Forum

#1 2015-04-24 23:18:27

quadratic polynomial

#2 2015-04-24 23:31:21

Re: quadratic polynomial

#3 2015-04-25 00:33:22

Re: quadratic polynomial

#4 2015-05-01 15:09:12

Re: quadratic polynomial

#5 2015-05-01 17:07:04

Re: quadratic polynomial

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Re: quadratic polynomial

#7 2015-05-02 03:54:25

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#8 2015-05-02 04:37:48

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#9 2015-05-02 05:09:49

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#10 2015-05-02 23:06:06

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Re: quadratic polynomial

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#13 2015-05-05 10:34:53

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#14 2015-05-05 10:57:52

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