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I have recently been researching about quadratic equations, and came across the following formula for finding the minimum $x$ and $y$ values from a "Bivariate (two equation) quadratic":

$$x = -(2BC-DE)/(4AB-E^2)$$ $$y = -(2AD-CE)/(4AB-E^2)$$

Where the variables are:

$Ax^2 + By^2 + Cx + Dy + Exy + F$

If I wanted to expand this to having $n$ variables (for example: for $n=3$ it would be $x,y,z$), what method would I use to create these formulas for finding the minimum values of all the variables?

Edit: I'm still quite new to this site, if there's any way I can improve my question please edit it or let me know!

user376343
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Rlz
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    You cancel the gradient of the function, which gives a square system of linear equations. This gives a stationary point, which can be maximum, minimum or saddle. –  May 02 '19 at 19:06
  • Could you please include this as an answer, I'm not sure what you mean by that – Rlz May 02 '19 at 19:14
  • "$\textrm{ I'm still quite new to this site}$": Yeah, $3$ years and $2$ months is not a long time especially when you are a Ocean Quahog – callculus42 May 02 '19 at 19:22

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Let's say that you have a function $F(x_0,x_1...x_n)$.

A critical point is any at which all of the derivatives $\displaystyle \frac{\partial F}{\partial x_0}....\frac{\partial F}{\partial x_n}=0$.

You would have a system of equations which you could then solve. Because it's a multivariable quadratic, all the derivatives are linear, and so you can use regular matrices to solve for the point of minimum.

The minimum would only occur at a point where all the second derivatives are negative.

This is still not guaranteed, but the set of points you solve for is where the minimum MAY lie. There may also be saddle points.

Saketh Malyala
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