Problem 20 - College Algebra

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. $$ x^{2}+2 x<0 $$

Step-by-Step Solution

Answer

The solution to the inequality \(x^2 +2x < 0\) is \((0, +\infty)\)

Step 1: Standard Form

First, note that the inequality is in standard form \(ax^2 + bx < 0\)

Step 2: Factoring

Next, factor the polynomial, which yields \(x(x + 2) < 0\)

Step 3: Finding critical points

The critical points are found by setting each factor equal to 0 and solving. So the critical points are \(x = 0\) and \(x = -2\)

Step 4: Setting up the Number Line and regions

Make a number line and plot these points. There are now three regions to consider: \(x < -2\), \(-2 < x < 0\), and \(x > 0\)

Step 5: Testing the intervals

Test a number in each region. If a region satisfies the inequality, it is a part of the solution set. For \(x < -2\), the inequality becomes negative and does not satisfy the inequality. For \(-2 < x < 0\), the inequality also becomes negative and does not satisfy the inequality. For \(x > 0\), \(x(x + 2)\) becomes positive and satisfies the inequality.

Step 6: Interval Notation

Finally, the solution in interval notation is \((0, +\infty)\)

Key Concepts

Factoring Polynomials

When you are faced with a polynomial inequality like the one in our example, the first step is to factor the polynomial. Factoring is the process of breaking down a complex expression into simpler components. These components, or factors, when multiplied together, yield the original expression. In our problem, the inequality starts as \(x^2 + 2x < 0\). To factor this, you look for terms that multiply to give you the original polynomial. Here, it is factored into \(x(x + 2)\). This is because multiplying \(x\) by \((x + 2)\) returns \(x^2 + 2x\).

Factoring makes it easier to analyze the behavior of the polynomial across different values of \(x\). It's an essential step because it reveals "critical points," which we will discuss next.

Critical Points

Critical points are the values of \(x\) where the polynomial equals zero. These points are important as they help divide the number line into sections where the inequality might change its truth value.

To find the critical points, you take each factor from the factored form and set it equal to zero. So, for our example \(x(x + 2) = 0\), the critical points are:

• \(x = 0\)
• \(x = -2\)

These critical points divide the number line into regions which can now be analyzed individually to determine where the inequality holds true.

Number Line

Plotting on a number line is a great way to visualize the critical points and determine the intervals to test. Once you've found your critical points \(x = 0\) and \(x = -2\), you place them on a number line. They divide the line into three regions:

• Region 1: \(x < -2\)
• Region 2: \(-2 < x < 0\)
• Region 3: \(x > 0\)

Testing each region by choosing a test point helps determine if the original inequality holds in that section. Checking each region, you look at the sign of the expression, not the exact value, to assess where the inequality is satisfied.

Interval Notation

Once you identify which regions satisfy the inequality using test points, you express these solutions in interval notation. Interval notation is a way of writing subsets of the real number line. It uses brackets to show the start and end of the interval with parentheses showing whether endpoints are included or excluded.

In our example, only \(x > 0\) satisfies the inequality, as the expression becomes positive in this region. Therefore, your solution is written in interval notation as:

• \((0, +\infty)\)

This indicates that all numbers greater than 0 satisfy the inequality, up to infinity, but not including 0 itself. Interval notation is concise and effective for representing a range of solutions.