Problem 22 - 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. $$ x^{2}+2 x<0 $$

Step-by-Step Solution

Answer

The solution set for the given quadratic inequality \(x^{2}+2 x<0\) is \((-\infty, -2) \cup (0, +\infty)\).

Step 1: Factorize the quadratic expression

Factorize the quadratic expression, \(x^{2}+2 x<0\), to find the roots. To do this, group the terms: \(x(x + 2) < 0\). This yields roots \(x=0\) and \(x=-2\)

Step 2: Determine the solution intervals

Now it's time to determine on which intervals the quadratic expression is negative. There are three intervals resulting from the roots: \(-\infty, -2\), \(-2, 0\), and from \(0, +\infty\).

Step 3: Test the intervals

Now pick a number from each interval and substitute it in the inequality to see if the inequality holds true. Numbers for test could be -3, -1, and 1. Substituting these values gives:From \(-\infty, -2\): Substituting -3 into \(x(x + 2) < 0\), we get 3 > 0, which is true.From \(-2, 0\): Substituting -1 into \(x(x + 2) < 0\), we get -1 > 0, which is false.From \(0, +\infty\): Substituting 1 into \(x(x + 2) < 0\), we get 3 > 0, which is true. Therefore, the solution intervals are \(-\infty, -2\) and \(0, +\infty\).

Step 4: Express the solution set in interval notation

Express the solution in interval notation. Solution set is: \((-\infty, -2) \cup (0, +\infty)\). This represents all real numbers less than -2, and all real numbers greater than 0.

Step 5: Graph the solution set on a real number line

Mark the points \(x=-2\) and \(x=0\) on the number line. Draw a line from \(-\infty\) to -2, and from 0 to \(\infty\). The area between -2 and 0 is not included because the inequality does not include equality.

Key Concepts

Factorization

Solving quadratic inequalities often begins with factorization. In this context, factorization means expressing a quadratic expression, such as \(x^2 + 2x \), in a form that helps us find its roots. The expression can be rewritten as \(x(x + 2)\). You have factored it into two simpler expressions (or factors). Each factor gives us a potential solution where the inequality is zero. Here, it gives the potential solutions \(x = 0\) and \(x = -2\). These points are crucial for the next steps, as they help to identify the intervals for further testing.

Interval Notation

Interval notation provides a compact way to express ranges of values that satisfy an inequality. After factorization shows the roots, the solution intervals between and around these roots are determined. For an inequality like \(x(x + 2) < 0\), you identify the intervals that make the product negative. Here, the intervals are \((-\infty, -2)\), \((-2, 0)\), and \((0, \infty)\).

• \((-\infty, -2)\): All real numbers less than \(-2\).
• \((-2, 0)\): All real numbers between \(-2\) and \(0\) (exclusive).
• \((0, \infty)\): All real numbers greater than \(0\).

By testing each interval, you find which one satisfies the original inequality.

Number Line Graphing

Graphing on a number line helps you visualize the solution set of the inequality. After finding the meaningful intervals, \((-\infty, -2)\) and \((0, \infty)\), you need to draw these on a number line. First, mark the points \(x = -2\) and \(x = 0\) with open circles. This indicates these points are not in the solution set as the inequality is strict (not "less than or equal to").
Then, shade the region to the left of \(x = -2\) and to the right of \(x = 0\). This graphic representation confirms visually which values satisfy the inequality, making it easier to understand the solution.

Solution Intervals

Solution intervals are derived after testing the values from the possible ranges exposed by factorization. By substituting test points such as \(-3\), \(-1\), and \(1\) into the inequality \(x(x + 2) < 0\), you'll know which intervals satisfy it.

• \(-3\) makes \(3 > 0\), indicating the interval \((-\infty, -2)\) works.
• \(-1\) does not satisfy the inequality, ruling out \((-2, 0)\).
• \(1\) makes an expression that matches the inequality's requirement, proving the interval \((0, \infty)\) is valid.

The solution then is expressed as \((-\infty, -2) \cup (0, \infty)\), indicating the set of x-values in these intervals satisfy the quadratic inequality.