# Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 2

### If $y=\frac{x^2+3}{x}$, find x as a function of y.

SOLUTION:

$y=\frac{x^2+3}{x}$

$xy=x^2+3$

$x^2-xy+3=0$

Solve for x using quadratic formula. We have $a=1,\:b=-y,\:and\:c=3$

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

$x=\frac{\cdot -\left(-y\right)\pm \sqrt{\left(-y\right)^2-4\left(1\right)\left(3\right)}}{2\left(1\right)}$

$x=\frac{y\pm \sqrt{y^2-12}}{2}$