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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.

]]>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.

]]>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

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

]]>( - 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

]]>It was quietly easy.

Thanks for helping me out.]]>

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.

]]>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

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

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

]]>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

]]>.

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