Problem 34 - Algebra and Trigonometry

Question

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ (5-x)^{2}\left(x-\frac{13}{2}\right)<0 $$

Step-by-Step Solution

Answer

The solution to the polynomial inequality \((5-x)^{2}\left(x-\frac{13}{2}\right)<0\) in interval notation is \((- \infty, 5) \cup (6.5, \infty)\).

Step 1: Factor Polynomial

The polynomial is already factored in the exercise which is \((5-x)^{2}\left(x-\frac{13}{2}\right) < 0\)

Step 2: Identify Critical Points

The polynomial will be zero at its critical points. These points are the solutions to the equation \((5-x)^{2}\left(x-\frac{13}{2}\right) = 0\). This gives the solutions \(x=5\) (from \(5-x=0\)) and \(x=\frac{13}{2}\) (from \(x-\frac{13}{2}=0\)). Thus, the critical points are \(5\) and \(\frac{13}{2}\) or \(6.5\).

Step 3: Test the Intervals

The critical points divide the number line into 3 regions: \(-\infty\) to \(5\), \(5\) to \(\frac{13}{2}\ = 6.5\), and \(6.5\) to \(\infty\). Pick a number from each of these intervals and substitute into the inequality. Choose \(x=4\), \(x=6\) and \(x=7\) for these tests. For \(x=4\), the inequality will be \((5-4)^2 * (4-\frac{13}{2}) > 0\), which is positive. For \(x=6\), the inequality will be \((5-6)^2 * (6-\frac{13}{2}) < 0\), which is negative. For \(x=7\), the inequality will be \((5-7)^2 * (7-\frac{13}{2}) > 0\), which is positive. Hence, the inequality holds true for \(-\infty < x < 5\) and \(6.5 < x < \infty\)

Step 4: Write in Interval Notation

The solution in interval notation is \(-\infty < x < 5\) and \(6.5 < x < \infty\). This interval can be written as \((- \infty, 5) \cup (6.5, \infty)\)