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#1 2007-09-12 09:24:42

tony123
Member
Registered: 2007-08-03
Posts: 228

solve in real numbers

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#2 2007-09-13 01:06:50

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: solve in real numbers

The only deduction I've made so far is that

.

That isn't a complete solution, there are other restrictions in there that I haven't worked out yet.
One exception: (0,0) is a correct solution that the above restriction doesn't apply to.


Why did the vector cross the road?
It wanted to be normal.

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#3 2007-09-13 02:01:28

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

Re: solve in real numbers

We have:

and

Using the second equation gives:

Call the above equation (1)

Equation (1) can also be written:

Substituting these values for x and its square into the first of the original equations gives:

Each term has y as a factor, so y = 0 is a solution, putting this value for y into equation (1) gives x = 0, therefore (x = 0,y= 0) is one solution of the original equations. Now lets divde through by y giving:

divide through by 2 giving:

and so

giving

Now square both sides giving:

Expand to give:

which is to say:

This is quadratic in y cubed, let z = y^3 giving:

Using the quadratic formula this gives:

therefore

or

and so

or

and so, finally:

or

There are two other solutions here for y of:

and

But due to the fractional powers of the negative quantities, these are not real.

So in summary we have found solutions when:

Which gives our solution set as (Using equation (1) to find x for each y):

Last edited by gnitsuk (2007-09-13 03:24:47)

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