I am solving some differential equations "solvable for $p$" and I came up with the following equation:

$$p^2 + p \left( \frac{y}{x} - \frac{x}{y} \right) - 1 = 0$$

I never came across such an equation before. Is this a quadratic equation and why? because as far as I know the form should be the following:

$$ax^2 + bx + c = 0$$

Where $a, b, c$ are constants but in this case it seems different.


The equation is a quadratic in variable $p$. To solve it, you can either complete the square or use the quadratic formula

Quadratic formula:

$$ p = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

where $a = 1$, $b = \frac{y}{x} -\frac{x}{y} = \frac{y^2-x^2}{xy}$, $c=-1$

You can then make the necessary substitution to obtain a solution for $p$

$$ p = \frac{1}{2} \left(\frac{x^2-y^2}{xy}\pm \sqrt{\left(\frac{y^2-x^2}{xy}\right)^2 + 4}\right)$$

$$ p = \frac{1}{2} \left(\frac{x^2-y^2}{xy} \pm \sqrt{\frac{y^4+2x^2y^2+x^4}{x^2y^2}}\right) $$

$$p = \frac{1}{2} \left( \frac{x^2-y^2}{xy} \pm \frac{x^2+y^2}{xy}\right) $$

$$ p = \frac{x}{y} \ \text{or} \ -\frac{y}{x}$$

where $y \ne 0$, $x \ne 0$

  • $\begingroup$ When you say, the equation is quadratic in p, instead of it is quadratic equation what does it mean? $\endgroup$ – user10213 Sep 26 '15 at 7:58
  • $\begingroup$ For example, $ax^2+bx +c = 0$ is a quadratic in unknown variable $x$. You need to be clear what the unknown variable is because the equation can be re-arranged to a quadratic in variables $x$ or $y$ for example. $\endgroup$ – John_dydx Sep 26 '15 at 8:03
  • $\begingroup$ Also, what happened to the 4 in the equation? $\endgroup$ – user10213 Sep 26 '15 at 8:03
  • $\begingroup$ @user10213, I've omitted a few steps-you need to expand the bracket and collect like terms, add the fraction etc. $\endgroup$ – John_dydx Sep 26 '15 at 8:04

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