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There's a formula for solving a Quadratic Equation that has $x$ as a variable; e.g:

$$2x^2 + 5x - 8x = 0$$

Step 1 (is to find delta/discriminant):

$\Delta = b^2 - 4ac$

Step 2 (is to find $x_1$ and $x_2$):

$x$ = $\dfrac{-(b) \pm \sqrt\Delta}{2a}$


My question is... Is it possible to solve an equation which has $x$ & $y$ variables, with the formula above? Or is there another formula for equations with $x$ & $y$ variables?

Example:

$$2x^2 + 5xy - 8y^2 = 0$$

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  • $\begingroup$ To solve an equation with two variables, we need two equations in the first place. With your example, you could solve for $x$ and your answers would have a $y$ in them, or solve for $y$ where the ansswers would have an $x$ in them. There is no way to solve this equation for $x$ and $y$ as it stands now $\endgroup$ – lioness99a Jun 22 '17 at 14:07
  • $\begingroup$ We could either treat it as a quadratic in $x$, where $a=2$, $b=5y$ and $c=-8y^2$, or a quadratic in $y$ where $a=-8$, $b=5x$ and $c=2x^2$ $\endgroup$ – lioness99a Jun 22 '17 at 14:08
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    $\begingroup$ An equation of the form $ax^2+by^2+cxy+dx+ey+f=0$ has in general infinitely many solutions. It describes a conic: a circle, an ellipse, a parabola, ... See en.wikipedia.org/wiki/Conic_section $\endgroup$ – mfl Jun 22 '17 at 14:31
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Going off of @lioness99a's comment, for your example you can set $a=2$, $b=5y$, and $c=-8y^2$ to get

\begin{align*} x &= \frac{-5y \pm \sqrt{(5y)^2 - 4\cdot 2 \cdot (-8y^2)}}{2\cdot 2} \\ &= \frac{-5y \pm \sqrt{89}|y|}{4} \\ &= \frac{1}{4} (-5 \pm \sqrt{89})y \end{align*}

or fix $x$ and treat $y$ as the variable.

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  • $\begingroup$ Because of the $\pm$, you could use just $y$ instead of $|y|$. and get $x = = \frac{-5 \pm \sqrt{89}}{4}y$ $\endgroup$ – steven gregory Jun 22 '17 at 15:22
  • $\begingroup$ thanks, I didn't notice that $\endgroup$ – Dando18 Jun 22 '17 at 15:23
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HINT: if $$y=0$$ and $$x=0$$ are not a solutions of your equation you can write $$2\frac{x}{y}+5-8\frac{y}{x}=0$$ Setting $t=x/y$ you will have $$2t+5-\frac{8}{t}=0$$ multiplying by $t$: $$2t^2+5t-8=0$$

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    $\begingroup$ How does this help find the values of $x$ and $y$? You can solve this for $t$ and then you are still left with an equation of the form $\frac xy=t$ to solve. Also, in the example, $(x,y)=(0,0)$ is a solution to the equation $\endgroup$ – lioness99a Jun 22 '17 at 14:11
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    $\begingroup$ if you have for example $$x/y=5$$ then is $$x=5y$$ and you can solve your equation for $y$ and then you have $x$ $\endgroup$ – Dr. Sonnhard Graubner Jun 22 '17 at 14:26

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