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Mathematics
Tag Archives: calculus solutions
Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 2
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PROBLEM:
If \displaystyle y=\frac{x^2+3}{x}, find x as a function of y.
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SOLUTION:
\begin{align*}
y & = \frac{x^2+3}{x} \\
xy & =x^2+3 \\
x^2-xy+3&=0
\end{align*}Solve for x using the quadratic formula. We have a=1,\:b=-y,\:\text{and}\:c=3
\begin{align*}
x & =\frac{-b\pm \sqrt{b^2-4ac}\:}{2a} \\
x & =\frac{ -\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} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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