Problem 15 - College Algebra

Question

Solve each quadratic inequality in Exercises \(1-28\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ 2 x^{2}+x<15 $$

Step-by-Step Solution

Answer

The solution is \((-∞, -3] ∪ [-3, 2.5)\). This is the representation of the solution on the number line in interval notation.

Step 1: Rearrange the inequality

Rewrite the inequality in the standard form of a quadratic equation: \( 2x^{2} + x - 15 < 0 \). This is done by subtracting 15 from both sides of the original given inequality.

Step 2: Factor the quadratic equation

Factor the quadratic equation: \( 2x^{2} + x - 15 = 0 \). This gives: \( (2x - 5)(x + 3) = 0 \).

Step 3: Identify and test critical points

Solve \( (2x - 5)(x + 3) = 0 \) for x. This gives the critical points \( x = 2.5 \) and \( x = -3 \). Divide the number line into intervals using these critical points and choose a test point from each interval: 1. Interval \((-∞, -3)\), test point -4.2. Interval \((-3, 2.5)\), test point 0.3. Interval \((2.5, ∞)\), test point 3. Substitute each test point into \( (2x - 5)(x + 3) \), If the result is less than 0, then the test point satisfies the inequality.

Step 4: Determine the solution set

The test points in intervals \((-∞, -3)\) and \((-3, 2.5)\) satisfy the inequality, hence the solution set is \((-∞, -3] ∪ [-3, 2.5)\). The test point in interval \((2.5, ∞)\) does not satisfy the inequality.

Step 5: Express in interval notation

Express the solution in interval notation as \((-∞, -3] ∪ [-3, 2.5)\).

Key Concepts

Solution Set

When solving quadratic inequalities, our goal is to find the range of values for which the inequality holds true. This range is known as the solution set. For the inequality \(2x^{2} + x - 15 < 0\), solving it involves determining the x-values that satisfy this condition.

To achieve this, we first rewrite the inequality in standard form and then factor it. By analyzing the factored expression, we can find the regions on the number line that satisfy the inequality. These regions collectively make up the solution set. For our example, the solution set is represented as \((-\infty, -3] \cup [-3, 2.5)\). This shows the x-values that make the inequality true.

Interval Notation

Interval notation is a way of representing a set of numbers along a number line. It's widely used in expressing the solution sets for inequalities. With interval notation, we use brackets and parentheses to denote which numbers are included or excluded.

• Parentheses \(()\): Denotes that an endpoint is not included.
• Square brackets \([]\): Denotes that an endpoint is included.

For our inequality \(2x^{2} + x - 15 < 0\), the solution set is written as \((-\infty, -3] \cup [-3, 2.5)\). This means all numbers between negative infinity and \(-3\) (including \(-3\)) and numbers between \(-3\) and \(2.5\) (excluding \(2.5\)) are solutions.

Factoring Quadratic Equations

Factoring quadratic equations is a key step in solving quadratic inequalities. It involves expressing the equation as a product of linear factors. For the quadratic equation \(2x^{2} + x - 15 = 0\), factoring is crucial to find the critical points and analyze intervals.

The equation is factored as \((2x - 5)(x + 3) = 0\). This means the quadratic breaks down into two factors that multiply to zero at specific x-values: \(x = 2.5\) and \(x = -3\). Factoring helps us easily spot these critical points and provides a structure to evaluate the original inequality.

Critical Points

Critical points in a quadratic inequality are the values of x where the factors equal zero. These points help divide the number line into different intervals, each potentially offering a different outcome for the inequality.

For the inequality \((2x - 5)(x + 3) < 0\), solving the equation \((2x - 5)(x + 3) = 0\) gives us the critical points \(x = 2.5\) and \(x = -3\).

These points effectively split the number line into three regions: \((-\infty, -3)\), \((-3, 2.5)\), and \((2.5, \infty)\). By choosing test points in each interval, we determine which intervals satisfy the inequality, ultimately forming the solution set.