Solve for X and Y

Question

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!

Answer 1

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.

Answer 2

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 23 }{ 17 } & 1 & 0 \\ - \frac{ 125 }{ 34 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 34 & 0 & 0 \\ 0 & \frac{ 98 }{ 17 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 23 }{ 17 } & - \frac{ 125 }{ 34 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 34 & 46 & - 125 \\ 46 & 68 & - 172 \\ - 125 & - 172 & 461 \\ \end{array} \right) $$

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