A peculiar equation

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DrZiro
Posts: 132
Joined: Mon Feb 09, 2009 3:51 pm UTC

A peculiar equation

Postby DrZiro » Tue Apr 08, 2014 1:51 pm UTC

A long time ago, I came up with an equation. It looks like this:

x^2 + y^2 + z^2 + 1 = x + (2y + √(23)z) / 3

where x, y and z are real.

The fun thing about it is that even though there are three variables, it can be completely solved. I just can't remember how.

So, how?

lalop
Posts: 210
Joined: Mon May 23, 2011 5:29 pm UTC

Re: A peculiar equation

Postby lalop » Tue Apr 08, 2014 2:13 pm UTC

Literal answer of "how", rather than actual solution:

Spoiler:
Completing the square, you'll end up with an equation of the form

(x-a)^2 + (y-b)^2 + (z-c)^2 = 0

where a,b,c are reals. This forces x-a = y-b = z-c = 0.

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jaap
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Re: A peculiar equation

Postby jaap » Tue Apr 08, 2014 2:18 pm UTC

Spoiler:
Complete the square for each of x, y, z separately, so that you only have quadratic terms and constant terms.

Spoiler:
It just so happens that the constant terms add up to zero...

Spoiler:
... leaving you with three squares that sum to zero.

Spoiler:
Therefore each of the squares is itself zero, which determines the value of each unknown.


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