Problem 69 - Algebra and Trigonometry

Question

Solve each inequality in Exercises \(65-70\) and graph the solution set on a real number line. $$ \frac{x^{2}-x-2}{x^{2}-4 x+3}>0 $$

Step-by-Step Solution

Answer

The solution to the inequality is \(x < -1\) or \(x>2\).

Step 1: Simplify the Inequality

First, simplify the inequality by factoring the numerator and the denominator.So, \(\frac{x^{2}-x-2}{x^{2}-4 x+3}>\) can be written as \(\frac{(x-2)(x+1)}{(x-1)(x-3)}>0\)

Step 2: Find the Critical Points

Critical points are the values of \(x\) at which the given equation equals zero or is undefined. For \(\frac{(x-2)(x+1)}{(x-1)(x-3)}=0\), we get \(x=2, -1\). The values of \(x\) where the equation is undefined are when \(x=1, 3\). So, the critical points are \(x=-1, 1, 2, 3\).

Step 3: Check the Intervals

Now, we need to test the sign of the expression on intervals determined by these critical points. Choose a test point from each interval and substitute that in the inequality.\nThe intervals and their signs will be: \nFor \(- \infty < x < -1\), choose \(x=-2\), the sign is +; \nfor \(-1 < x < 1\), choose \(x=0\), the sign is -; \nfor \(1 < x < 2\), choose \(x=1.5\), the sign is -; \nfor \(2 < x < 3\), choose \(x=2.5\), the sign is +, \nand for \(3 < x < \infty\), choose \(x=4\), the sign is +.

Step 4: Drawing the Solution on the Number Line

Mark the critical points \(x=-1, 1, 2, 3\) on the number line. The inequality is \(>\) 0. So we are looking for the intervals where the sign is positive. These are the intervals \(- \infty < x < -1\) and \(2 < x < \infty\). Shade these intervals on the number line. However, since there is no “equal to” in the inequality, we put open circles at the critical points.