Problem 40 - Precalculus Mathematics for Calculus

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$x^{2}+5 x+6>0$$

Step-by-Step Solution

Answer

Solution: \((-\infty, -3) \cup (-2, \infty)\).

Step 1: Factor the Quadratic Expression

The given inequality is \(x^2 + 5x + 6 > 0\). To solve the inequality, we first factor the quadratic expression. Let's find two numbers that multiply to 6 and add to 5. These numbers are 2 and 3. So, the factorization is \((x+2)(x+3)\).

Step 2: Identify Critical Points

The expression \((x+2)(x+3) > 0\) defines the inequality. The critical points occur where each factor is equal to zero, i.e., \(x+2=0\) or \(x+3=0\). Solving these, we get the critical points \(x=-2\) and \(x=-3\).

Step 3: Determine Intervals and Test Sign

The critical points divide the number line into three intervals: \((-\infty, -3)\), \((-3, -2)\), and \((-2, \infty)\). We need to test each interval to determine where the product \((x+2)(x+3)\) is positive.- Select a test point from each interval: - For \((-\infty, -3)\), test \(x = -4\); calculate \((-4+2)(-4+3) = (-2)(-1) = 2\) (positive). - For \((-3, -2)\), test \(x = -2.5\); calculate \((-2.5+2)(-2.5+3) = (-0.5)(0.5) = -0.25\) (negative). - For \((-2, \infty)\), test \(x = 0\); calculate \((0+2)(0+3) = (2)(3) = 6\) (positive).

Step 4: Write Solution Using Interval Notation

From the signs of the test points, \((x+2)(x+3) > 0\) is true on the intervals \((-\infty, -3)\) and \((-2, \infty)\). The solution in interval notation is \((-\infty, -3) \cup (-2, \infty)\).

Step 5: Graph the Solution

To graph the solution:- Draw a number line.- Mark the critical points \(-3\) and \(-2\) with open circles (since the inequality is strict \(>\)), indicating that these points are not included in the solution.- Shade the region corresponding to \((-\infty, -3)\) and \((-2, \infty)\).

Key Concepts

Factorization

Factorization is an essential step in solving nonlinear inequalities. It involves breaking down a complex expression into products of simpler factors. By identifying factors, you can simplify problems and find solutions more efficiently.
To factor a quadratic expression such as the given example, \(x^2 + 5x + 6 > 0\), you look for two numbers that multiply to the constant term (6) and add to the linear coefficient (5). In this case, the numbers are 2 and 3. Thus, the expression becomes \((x+2)(x+3)\).
Factorization helps in determining where the expression changes sign. This will aid in solving the inequality by making it easier to identify intervals where the expression is positive or negative. Always double-check your factorization to ensure accuracy.

Interval Notation

Interval notation is a concise way to express solutions to inequalities. It uses brackets and parentheses to describe intervals of numbers on the number line.
For this problem, the solution to the inequality \((x+2)(x+3) > 0\) was found to be on the intervals \((-fty, -3)\) and \((-2, fty)\).

• Parentheses \(()\) are used to indicate that a number is not included in the interval, which is appropriate when dealing with strict inequalities (using \(<\) or \(>\)).
• Brackets \([]\) would be used if the endpoints were to be included, common in non-strict inequalities (\(\leq\) or \(\geq\)).

Here, since the inequality is strict, we use open parentheses around the numbers -3 and -2, indicating these points are excluded.

Critical Points

Critical points arise in inequalities where each factor of the factored expression equals zero.
For the inequality \((x+2)(x+3) > 0\), critical points are found by solving the equations \(x+2=0\) and \(x+3=0\). This gives the critical points \(x=-2\) and \(x=-3\).
These points are important since they signify where the sign of the expression might change. Critical points help divide the number line into distinct intervals, allowing one to test each interval's sign individually.

• Once critical points are identified, you can easily sketch intervals on the number line.
• Each interval test will determine if that segment satisfies the original inequality.

Remember, critical points themselves are generally not included in the solution unless specified by the inequality's conditions.

Number Line

A number line visually represents solutions and intervals. When dealing with inequalities, it's a useful tool to help understand the distribution of solutions and marks different regions dictated by critical points.
In this exercise, the number line is divided into intervals based on the critical points, \(x = -3\) and \(x = -2\).

To use a number line effectively:

• Mark the critical points with open circles if they are not included in the solution (as indicated by a strict inequality).
• Shade the segments of the number line corresponding to positive test results for their respective intervals.

Visually interpreting the solution on the number line helps to reinforce the understanding of the problem's interval notation, offering a clear view of where solutions lie. It's a key step in solving and graphing inequalities.