Solving Rational Inequalities Analytically

Exercise 4.

Find the solutions of the inequality

\begin{displaymath}\frac{x^2-4}{x^2-1}\leq \frac{x}{x+3}.\end{displaymath}

Answer.

Start by moving the term on the right to the left:

\begin{displaymath}\frac{x^2-4}{x^2-1}-\frac{x}{x+3}\leq 0.\end{displaymath}

Combine the terms on the left and simplify:

\begin{displaymath}\frac{x^2-4}{x^2-1}-\frac{x}{x+3}=\frac{(x^2-4)(x+3)-x(x^2-1)}{(x^2-1)(x+3)}=\frac{3(x^2-x-4)}{(x^2-1)(x+3)}.\end{displaymath}

By now the inequality looks like

\begin{displaymath}\frac{3(x^2-x-4)}{(x^2-1)(x+3)}\leq 0.\end{displaymath}

By using the quadratic formula, we find that the equation x2-x-4=0 has as its roots

\begin{displaymath}x=\frac{1-\sqrt{17}}{2}\approx -1.56155
\mbox{ and }x=\frac{1+\sqrt{17}}{2}\approx 2.56155.\end{displaymath}

The denominator is zero at $x=\pm 1$ and x=-3.

The solutions are all points in the set

\begin{displaymath}(-\infty,-3)\cup \left[\frac{1-\sqrt{17}}{2},-1\right)\cup \left(1,\frac{1+\sqrt{17}}{2}\right].\end{displaymath}

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Helmut Knaust
1998-06-16

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