Quadrilateral problem - Page 2

Kees · 2007-01-31, 12:24

My avatar is a very grumpy cat called Heinz, definitely not a female cat.
As there is no Heinz smiley (well, probably copyright reasons) I had to use another one.

Wacky · 2007-01-31, 13:38

Quote:

Originally Posted by mfgoode

Take a look at the bottom line of your post 2 and you will know who hid your Half baked solution. As is common in this thread once again it was itself half baked without the tip off like ATH's

Mally

Mally,
Your comments are uncalled for.
Anyone who would like to see exactly how I hid his response need only "quote" the message in question and look at the resulting text in the edit window.

mfgoode · 2007-01-31, 15:47

Quote:

Originally Posted by S485122

This is not correct : it should be at least one root is a real number. For instance :
x³-6*x²+11*x-6 =0 has three real solutions, i.e; 1, 2 & 3
x³-4*x²+5*x-2=0 has two real solutions, i.e; 1 & 2 (2 being a double solution)
x³-3*x²+3*x-1=0 has one real solution, i.e; 1 (it is a triple solution)

Of course if the coefficients are randomly chosen reals, most of the time there will be only one real solution and two more can be found by using imaginary numbers.

Jacob

Not quite so Jacob.
I'm afraid you are mixing up real and irrational numbers with rational numbers.
The real numbers consist of both rational and irrational, and complex numbers.

My quote is from the text book 'Higher Algebra' by Hall and Knight 2001 edition updated thru several editions and first published in 1887. That was about the time when these differences were rigorously defined and still stands good at the present time.

Moreover most irrational numbers are transcendental, unlike the rationals, which like algebraic, are countable.

However, thank you for bringing this up as its a good observation on your part.

Not being content with my text book I had to get on the Net and this is what I found.

http://en.wikipedia.org/wiki/Real_number

If you still disagree I am open to discussion as numbers have always fascinated me from childhood and I am willing to learn from you.

Mally

mfgoode · 2007-01-31, 15:55

Quote:

Originally Posted by Wacky

Mally,
Your comments are uncalled for.
Anyone who would like to see exactly how I hid his response need only "quote" the message in question and look at the resulting text in the edit window.

Well I often do that myself when Im in doubt.

But a straight question deserves a straight answer not an anticipated revelation.

It is only for the glory of God who has the right to hide things :not us mortals

Mally

wblipp · 2007-01-31, 16:33

Quote:

Originally Posted by mfgoode

If you still disagree I am open to discussion

If every cubic equation has a rational root, what is the rational root of

x³-2 = 0

Wacky · 2007-01-31, 16:49

Quote:

Originally Posted by mfgoode

But a straight question deserves a straight answer

Where was there a question? In English, such queries customarily end with a particular punctuation.

I read Kees statement as an apology that he had felt that it was appropriate to obscure his response, but that he had failed to do so.

As a moderator, I obliged him for the benefit of the other readers.

When, at message #5, he made a request for details, the answer was directly given. Since both inquiry and answer occurred while I was asleep, I saw no reason to elaborate.

S485122 · 2007-01-31, 17:19

Quote:

Originally Posted by mfgoode

I'm afraid you are mixing up real and irrational numbers with rational numbers.
The real numbers consist of both rational and irrational, and complex numbers.

In mathematics, the real numbers may be described informally as a number that can be given by an infinite decimal representation, such as 2.4871773339…. The real numbers include both rational numbers, such as 42 and −23/129, and irrational numbers, such as π and the square root of 2, and can be represented as points on an infinitely long number line. No mention of complex numbers !
As pointed out by Wblipp x³=2 has no rational solution, but it has a irrational (no ratios) solution meaning the solution is part of the real numbers.

Quote:

Originally Posted by mfgoode

My quote is from the text book 'Higher Algebra' by Hall and Knight 2001 edition updated thru several editions and first published in 1887. That was about the time when these differences were rigorously defined and still stands good at the present time.

I'm afraid you misquoted.

Quote:

Originally Posted by mfgoode

... most irrational numbers are transcendental, unlike the rationals, which <snip> are countable.

I fully agree.

Quote:

Originally Posted by mfgoode

Not being content with my text book I had to get on the Net and this is what I found http://en.wikipedia.org/wiki/Real_number

And that is where I found the definition I gave above ;-)

Jacob

philmoore · 2007-01-31, 20:27

The discriminant will tell whether there are three real roots or one real and two complex roots. In this case, the fact that the arithmetic mean of three positive numbers (at least two of which are distinct) is greater than the geometric mean implies that the discriminant is negative and all three roots are real. Now using the fact that the three roots must add up to zero, we can show that two roots are negative and that therefore the desired solution is the positive root.

Now the problem is that even though all three roots are real, the cubic formula expresses them in terms of complex numbers. We can take a cube root of a complex number by writing it in polar form. This gives us the solution (I am having trouble getting the spoiler tags to work with tex markup):

$d =\frac {10\sqrt6}{3}\cos(\frac{1}{3}\cos^{-1}\sqrt{\frac{486}{625}})$

Or more generally:

$d =2\sqrt{\frac{a^2+b^2+c^2}{3}}\cos\Big(\frac{1}{3}\cos^{-1}\sqrt{\frac{27a^2b^2c^2}{(a^2+b^2+c^2)^3}}\Big)$

Can anyone see any particular significance to the angle that is being trisected in this formula?

Kees · 2007-02-01, 09:25

To answer Wacky, it was indeed an apology and I am grateful for the spoilertags placed over my solution.

I see no reason why trisection should come into play in this problem or the solution.

mfgoode · 2007-02-01, 18:14

Quote:

Originally Posted by wblipp

If every cubic equation has a rational root, what is the rational root of

x³-2 = 0

:surprised

Well wblipp you've got me by a googly. I have to admit the roots are irrational so I can safely say that the roots are such that at least one is real as has been rightly pointed out by S485122.

However Hall and Night have an answer for this as roots have to be taken by pairs yz so the result is a rational, so they say, but I'm not sure right now.
I will quote them tomorrow as its too late now and the quote is quite lengthy.
I understand that if you take them separately one should get 9 roots for a cubic but this is not the case and the roots have to be paired to get only 3 as per the theory of equations. But don't take my word for it and lets wait and see.

Mally

mfgoode · 2007-02-01, 18:33

Quote:

Originally Posted by Kees

To answer Wacky, it was indeed an apology and I am ...

I see no reason why trisection should come into play in this problem or the solution.

I presume that to get the roots of a cubic is as good as trisecting an angle.

This is the argument and proof to the assertion by the Greeks of antiquity that an angle (not all but most) cannot be trisected by compass and unmarked ruler only. I have this proof in my library but will have to locate it to post it here. Well we can also get it form the Net.

http://mathworld.wolfram.com/AngleTrisection.html

This partly answers Philmoore's question but I would like to know his reason and thanks to your post phil as it was very enlightening.

Mally

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2007-01-31, 12:24	#12
Kees Dec 2005 2²·7² Posts	My avatar is a very grumpy cat called Heinz, definitely not a female cat. As there is no Heinz smiley (well, probably copyright reasons) I had to use another one.

2007-01-31, 20:27	#19
philmoore "Phil" Sep 2002 Tracktown, U.S.A. 3·373 Posts	The discriminant will tell whether there are three real roots or one real and two complex roots. In this case, the fact that the arithmetic mean of three positive numbers (at least two of which are distinct) is greater than the geometric mean implies that the discriminant is negative and all three roots are real. Now using the fact that the three roots must add up to zero, we can show that two roots are negative and that therefore the desired solution is the positive root. Now the problem is that even though all three roots are real, the cubic formula expresses them in terms of complex numbers. We can take a cube root of a complex number by writing it in polar form. This gives us the solution (I am having trouble getting the spoiler tags to work with tex markup): $d =\frac {10\sqrt6}{3}\cos(\frac{1}{3}\cos^{-1}\sqrt{\frac{486}{625}})$ Or more generally: $d =2\sqrt{\frac{a^2+b^2+c^2}{3}}\cos\Big(\frac{1}{3}\cos^{-1}\sqrt{\frac{27a^2b^2c^2}{(a^2+b^2+c^2)^3}}\Big)$ Can anyone see any particular significance to the angle that is being trisected in this formula?

2007-02-01, 09:25	#20
Kees Dec 2005 2²×7² Posts	To answer Wacky, it was indeed an apology and I am grateful for the spoilertags placed over my solution. I see no reason why trisection should come into play in this problem or the solution.