<![CDATA[Math Is Fun Forum / quadratic polynomial]]> 2015-05-08T08:11:31Z FluxBB http://www.mathisfunforum.com/viewtopic.php?id=22160 <![CDATA[Re: quadratic polynomial]]> http://www.mathisfunforum.com/viewtopic.php?id=22191

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http://www.mathisfunforum.com/profile.php?id=207651 2015-05-08T08:11:31Z http://www.mathisfunforum.com/viewtopic.php?pid=358650#p358650

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.

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http://www.mathisfunforum.com/profile.php?id=207651 2015-05-08T03:18:10Z http://www.mathisfunforum.com/viewtopic.php?pid=358634#p358634
<![CDATA[Re: quadratic polynomial]]> 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.

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http://www.mathisfunforum.com/profile.php?id=207651 2015-05-08T03:06:38Z http://www.mathisfunforum.com/viewtopic.php?pid=358633#p358633
<![CDATA[Re: quadratic polynomial]]> 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.

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http://www.mathisfunforum.com/profile.php?id=33790 2015-05-08T00:23:13Z http://www.mathisfunforum.com/viewtopic.php?pid=358626#p358626
<![CDATA[Re: quadratic polynomial]]> 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

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http://www.mathisfunforum.com/profile.php?id=67694 2015-05-07T22:23:12Z http://www.mathisfunforum.com/viewtopic.php?pid=358623#p358623
<![CDATA[Re: quadratic polynomial]]> Hi;

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

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http://www.mathisfunforum.com/profile.php?id=33790 2015-05-07T21:29:58Z http://www.mathisfunforum.com/viewtopic.php?pid=358621#p358621
<![CDATA[Re: quadratic polynomial]]> 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

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http://www.mathisfunforum.com/profile.php?id=67694 2015-05-07T15:32:11Z http://www.mathisfunforum.com/viewtopic.php?pid=358613#p358613
<![CDATA[Re: quadratic polynomial]]> well I came up with the same solution yesterday as olinguito.
It was quietly easy.
Thanks for helping me out.

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http://www.mathisfunforum.com/profile.php?id=198409 2015-05-07T13:15:53Z http://www.mathisfunforum.com/viewtopic.php?pid=358612#p358612
<![CDATA[Re: quadratic polynomial]]> 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.

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http://www.mathisfunforum.com/profile.php?id=33790 2015-05-06T19:18:55Z http://www.mathisfunforum.com/viewtopic.php?pid=358591#p358591
<![CDATA[Re: quadratic polynomial]]> 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

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http://www.mathisfunforum.com/profile.php?id=67694 2015-05-06T09:14:24Z http://www.mathisfunforum.com/viewtopic.php?pid=358575#p358575
<![CDATA[Re: quadratic polynomial]]> PS:

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

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http://www.mathisfunforum.com/profile.php?id=207651 2015-05-05T22:57:52Z http://www.mathisfunforum.com/viewtopic.php?pid=358537#p358537
<![CDATA[Re: quadratic polynomial]]> 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

In other words your <cube root c> minus <cube root a> must be equal to 1 – but as you can see they are not. ]]>
http://www.mathisfunforum.com/profile.php?id=207651 2015-05-05T22:34:53Z http://www.mathisfunforum.com/viewtopic.php?pid=358535#p358535
<![CDATA[Re: quadratic polynomial]]> 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

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http://www.mathisfunforum.com/profile.php?id=67694 2015-05-05T12:08:47Z http://www.mathisfunforum.com/viewtopic.php?pid=358454#p358454
<![CDATA[Re: quadratic polynomial]]> niharika_kumar wrote:

.

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http://www.mathisfunforum.com/profile.php?id=207651 2015-05-03T11:18:22Z http://www.mathisfunforum.com/viewtopic.php?pid=358358#p358358
<![CDATA[Re: quadratic polynomial]]> bob bundy wrote:

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

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http://www.mathisfunforum.com/profile.php?id=207651 2015-05-03T11:06:06Z http://www.mathisfunforum.com/viewtopic.php?pid=358357#p358357