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I have the following equation:

$0 = 34x^2+92xy+68y^2−250x−344y+461$

I cannot find any way to get the values of both x and y from this equation, any help would be much appreciated, especially a step by step solution.

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

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    $\begingroup$ One way to improve the post would be to present your work. Also, what do we know about $x$ and $y$? Complex, real, integer...? This equation surely has a continuum of complex roots. You can view it as a quadratic equation in $x$ and solve in terms of $y$ as a parameter. $\endgroup$ – A. Pongrácz May 2 at 16:07
  • $\begingroup$ They are all real numbers. Would a quadratic equation work with 10 or more variables? $\endgroup$ – Ruler Of The World May 2 at 16:09
  • $\begingroup$ You are changing your post in the wrong direction. Without showing any of your own efforts, you should definitely not ask for a step-by-step solution. As they say in shark tank, "I'm out". $\endgroup$ – A. Pongrácz May 2 at 16:11
  • $\begingroup$ Of course, my apologies. I'll keep the question the same and open another one if necessary. Let's focus on just two variables $\endgroup$ – Ruler Of The World May 2 at 16:14
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Solving for $x$ with the usual formula,

$$x=\frac{-(92y-250)\pm\sqrt{(92y-250)^2-4\cdot34\cdot(68y^2-344y+461)}}{68}.$$

After simplification, the discriminant is

$$-196(2y-1)^2,$$ which is only non-negative when $$y=\frac12.$$

$x$ follows.

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  • $\begingroup$ Great answer! What is the formula you used to do this? $\endgroup$ – Ruler Of The World May 2 at 16:20
  • $\begingroup$ @RulerOfTheWorld: as said in the comments. $\endgroup$ – Yves Daoust May 2 at 16:21
  • $\begingroup$ As far as I know, the quadratic formula is: x = (-b ± √(b²-4ac))/2a, however what you have used doesn't seem to be the same $\endgroup$ – Ruler Of The World May 2 at 16:26
  • $\begingroup$ @RulerOfTheWorld: it is exactly that one. What differences do you see ? $\endgroup$ – Yves Daoust May 2 at 17:33
  • $\begingroup$ Sorry for the late reply. I'm not sure where you got the values of a b and c from for the formula, I thought a=34 (from 34x^2), b=-250 (from -250x) and c=461 (from +461) $\endgroup$ – Ruler Of The World May 2 at 18:24
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$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 23 }{ 17 } & 1 & 0 \\ - \frac{ 125 }{ 34 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 34 & 0 & 0 \\ 0 & \frac{ 98 }{ 17 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 23 }{ 17 } & - \frac{ 125 }{ 34 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 34 & 46 & - 125 \\ 46 & 68 & - 172 \\ - 125 & - 172 & 461 \\ \end{array} \right) $$

So $$ x + \frac{ 23 }{ 17 } y - \frac{ 125 }{ 34 } = 0 $$ $$ y - \frac{ 1 }{ 2 } = 0 $$ because 34 times the square of the first one plus $ \frac{ 98 }{ 17 }$ times the square of the second must be zero.

The relationship between the matrices $D,H$ is called congruence

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