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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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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
orand so
orand so, finally:
orThere 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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