A peculiar equation

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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?

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Re: A peculiar equation

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

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

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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Re: A peculiar equation

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

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

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

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

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

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