Problem 78 - Algebra for College Students

Question

Solve each inequality. State the solution set using interval notation when possible. \((x-1)(x+2)(2 x-5)<0\)

Step-by-Step Solution

Answer

(\( -\infty, -2)\cup(1, \frac{5}{2})\).

Step 1 - Find the critical points

Set each factor equal to zero to find the critical points. \(x-1=0\), \(x+2=0\), and \(2x-5=0\). Solving these gives us: \(x=1\), \(x=-2\), and \(x=\frac{5}{2}\).

Step 2 - Determine the intervals

The critical points divide the number line into four intervals: \((-\infty, -2)\), \((-2, 1)\), \((1, \frac{5}{2})\), and \((\frac{5}{2}, \infty)\).

Step 3 - Test a point in each interval

Choose a test point from each interval and substitute it into the inequality \((x-1)(x+2)(2x-5)<0\):- Interval \((-\infty, -2)\): Test point \(x=-3\)\((-3-1)(-3+2)(2(-3)-5)=(-4)(-1)(-11) = -44\) (negative)- Interval \((-2, 1)\): Test point \(x=0\)\((0-1)(0+2)(2(0)-5)=(-1)(2)(-5) = 10\) (positive)- Interval \((1, \frac{5}{2})\): Test point \(x=2\)\((2-1)(2+2)(2(2)-5)=(1)(4)(-1) = -4\) (negative)- Interval \((\frac{5}{2}, \infty)\): Test point \(x=3\)\((3-1)(3+2)(2(3)-5)=(2)(5)(1) = 10\) (positive)

Step 4 - Determine the solution set

The solution set includes the intervals where the test point gives a negative result, as we need the expression to be less than zero. Therefore, the solution is the union of the intervals \((-\infty, -2)\) and \((1, \frac{5}{2})\).

Step 5 - State the solution in interval notation

Combine the intervals where the inequality holds true. The solution set in interval notation is: \((-\infty, -2)\cup(1, \frac{5}{2})\).

Key Concepts

critical points

To solve inequalities involving polynomials, we first need to find the critical points. Critical points are the values of the variable that make the expression equal to zero. These points help us understand where the inequality might change from positive to negative or vice versa.

In our example, the inequality is \( (x-1)(x+2)(2x-5)<0 \). We set each factor equal to zero and solve for x:

• \( x-1=0 \) results in x=1
• \( x+2=0 \) results in x=-2
• \( 2x-5=0 \) results in x=\(\frac{5}{2} \)

These values (-2, 1, \(\frac{5}{2} \)) are the critical points. They divide the number line into different intervals to analyze the inequality.

interval notation

Interval notation is a way to describe the set of solutions for an inequality. It uses parentheses and brackets to show which parts of the number line are included in the solution.

In our example, the critical points -2, 1, and \(\frac{5}{2} \) divide the number line into the following intervals:

• \((-\to -2) \)
• \((-2, 1) \)
• \((1, \frac{5}{2} \))
• \((\frac{5}{2}, \to ) \)

We will test each interval to see where the inequality \( (x-1)(x+2)(2x-5)<0 \) holds true.

test points

To determine where the inequality holds true, we use test points from each interval. By substituting these points into the original inequality, we can see whether the expression is positive or negative.

Here are the test points:

• For the interval \((-\to, -2)\), we use x=-3.
• For the interval \((-2, 1)\), we use x=0.
• For the interval \((1, \frac{5}{2})\), we use x=2.
• For the interval \((\frac{5}{2}, \to)\), we use x=3.

When substituted into the inequality \( (x-1)(x+2)(2x-5)<0 \):

• For x=-3: \( (-3-1)(-3+2)(2(-3)-5) = -44 \) (negative).
• For x=0: \( (0-1)(0+2)(2(0)-5) = 10 \) (positive).
• For x=2: \( (2-1)(2+2)(2(2)-5) = -4 \) (negative).
• For x=3: \( (3-1)(3+2)(2(3)-5) = 10 \) (positive).

This tells us where the expression is negative and thus where the inequality holds true.

solution set

The solution set is where the inequality holds true based on our test points. Since the inequality is \( < 0 \), we are looking for the intervals where the expression is negative.

From our test points:

• Interval \((-\to, -2)\) is negative.
• Interval \((1, \frac{5}{2})\) is negative.

Therefore, the solution set in interval notation is:

\((-\to, -2) \cup (1, \frac{5}{2})\) This shows that any x-value in these intervals will satisfy the original inequality.