Problem 12 - Beginning and Intermediate Algebra

Question

Solve each quadratic inequality. Graph the solution set and write the solution in interval notation. $$v^{2}+10 v<-16$$

Step-by-Step Solution

Answer

In short, for the quadratic inequality \(v^2 + 10v + 16 > 0\), the solution set is \((-{\infty}, -8) \cup (-2, \infty)\).

Step 1: Arranging the inequality in standard form

Move -16 to the left side: \(v^2 + 10v + 16 > 0\)

Step 2: Factoring the quadratic expression

Factor the left-hand side expression: \((v + 2)(v + 8) > 0\)

Step 3: Find the critical points

Set each factor equal to zero and solve for the roots: \(v + 2 = 0 \Rightarrow v = -2\) \(v + 8 = 0 \Rightarrow v = -8\) These roots are the critical points.

Step 4: Create a number line using the critical points

Place the critical points -2 and -8 on a number line.

Step 5: Test the intervals to determine their signs

Choose a value from each interval formed by the critical points and evaluate the inequality: Interval 1: \(v < -8\) (Choose v=-9) \((-7)(-1) > 0\) True Interval 2: \(-8 < v < -2\) (Choose v=-5) \((-3)(3) > 0\) False Interval 3: \(v > -2\) (Choose v=0) \((2)(8) > 0\) True

Step 6: Find the solution set and interval notation

The solution set consists of the intervals where the inequality is true: \(-9 < v < -8\) \(v > -2\) In interval notation, the solution set can be written as: \[(-\infty, -8) \cup (-2, \infty)\]

Key Concepts

Interval Notation

Interval notation is a method of describing sets of numbers on the number line. It is particularly useful in representing the solution sets of inequalities. This notation uses intervals to describe where a variable satisfies a given condition.
When dealing with quadratic inequalities like the one in our exercise, we need to identify ranges of values that satisfy the condition. In interval notation, square brackets

• For example, consider the inequality solution: \((-\infty, -8) \cup (-2, \infty)\)

Here, the brackets indicate that the numbers -8 and -2 are not included in the solution set because the inequality is strict ("greater than but not equal to" in this case). Understanding and using interval notation is key to accurately stating the solution to the inequality.

Factoring Quadratics

Factoring quadratics is a crucial step in solving quadratic inequalities. It involves breaking down a quadratic expression into a product of simpler expressions. This process allows us to find critical points, which in turn help in determining the solution set of the inequality.
In our exercise, we took the quadratic equation \(v^2 + 10v + 16 > 0\) and factored it into two binomials:

• \((v + 2)(v + 8) > 0\)

This tells us that the product of these two factors must be greater than zero. Successfully factoring the quadratic gives us the expressions we set equal to zero to find critical points. Overall, mastering the skills of factoring quadratics is essential for solving quadratic inequalities efficiently.

Solution Set

The solution set of a quadratic inequality is the set of all values that satisfy the inequality. After factoring the quadratic and finding the critical points, we test the regions those points create on the number line to determine the sign of the inequality in each region.
For example, with critical points at -8 and -2, the number line is divided into three intervals:

• \(v < -8\)
• \(-8 < v < -2\)
• \(v > -2\)

By testing a point from each interval in the original inequality, we determine which intervals satisfy the inequality. Here, the intervals \((-\infty, -8)\) and \((-2, \infty)\) satisfy the inequality, meaning these parts of the number line are the solution set.
Finding the solution set requires careful testing and validation, ensuring the inequality holds true within the intervals.

Critical Points

Critical points are the values that make the quadratic expression equal to zero when solving a quadratic inequality. These points are found by setting each factor of the factored quadratic equation to zero and solving for the variable.
In our scenario, the equation \((v + 2)(v + 8) > 0\) led us to the critical points at \(v = -2\) and \(v = -8\). These critical points divide the number line into distinct intervals we must test.

Critical points help determine where the quadratic changes from positive to negative or vice versa. Testing the sign of the quadratic in the intervals defined by the critical points helps establish the solution set for the inequality. This step is vital in identifying the valid segments of the number line for the solution set.