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I need to simplify the following equation, by completing the square: $\frac{\left(x-z\right)^{2}}{2(u-s)}+\frac{(z-y)^{2}}{2(t-u)}$

As $\displaystyle\frac{\left(x-y\right)^{2}}{2(t-s)}+C$.

How can I do this? I can't seem to be able to deal with the fractions effectively when completing this square...

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    Yes Sivaram - that is correct. And I would have done that, but the professor has explicitly said to calculate it by direct integration - to my demise. I basically end up with two lines of a letter-salad and can't get anywhere. By the way, I don't have to turn this in since it was due last week, but it is annoying the hell out of me.2011-02-14

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The only way to have $\frac{\left(x-z\right)^{2}}{2(u-s)}+\frac{(z-y)^{2}}{2(t-u)}$ equal to $\frac{\left(x-y\right)^{2}}{2(t-s)}+C$ is to let $C=\frac{(-s y+s z+t x-t z-u x+u y)^2}{2 (s-t) (s-u) (t-u)}.$ I'm not sure this is completing a square, though.

For extra fun, the numerator in $C$ is exactly $\left|\begin{array}{ccc}1&1&1\\s&t&u\\x&y&z\end{array}\right|^2,$ while the denominator is $-2\left|\begin{array}{ccc}1&1&1\\s&t&u\\s^2&t^2&u^2\end{array}\right|.$

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    Thanks. This is what I had - but I have to integrate the exponential function to the power of this, and I really didn't want to... :/2011-02-14