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


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


Solution:

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

\displaystyle xy=x^2+3

\displaystyle x^2-xy+3=0

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

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

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

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


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