Problem 52 - Algebra and Trigonometry

Question

\(33-66\) . Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ \frac{2 x+6}{x-2}<0 $$

Step-by-Step Solution

Answer

The solution is \((-3, 2)\), not including \(-3\) and \(2\).

Step 1: Identify Critical Points

To solve the inequality \( \frac{2x+6}{x-2}<0 \), first determine where the expression is equal to zero and where it is undefined. Set the numerator \(2x+6 = 0\) and the denominator \(x-2 = 0\) to find these points. Solving \(2x+6=0\), we get \(x=-3\). Solving \(x-2=0\), we get \(x=2\). These are the critical points: \(x=-3\) and \(x=2\).

Step 2: Solve the Inequality on Intervals

The critical points \(-3\) and \(2\) divide the number line into three intervals: \((-\infty, -3)\), \((-3, 2)\), and \((2, \infty)\). Test each interval in the inequality \(\frac{2x+6}{x-2}<0\) to determine where the inequality holds.- For \((-\infty, -3)\), pick \(x=-4\); the expression \(\frac{2(-4)+6}{-4-2} = \frac{-8+6}{-6} = \frac{-2}{-6}>0\).- For \((-3, 2)\), pick \(x=0\); the expression \(\frac{2(0)+6}{0-2} = \frac{6}{-2} = -3<0\).- For \((2, \infty)\), pick \(x=3\); the expression \(\frac{2(3)+6}{3-2} = \frac{6+6}{1} = 12>0\).

Step 3: Write Down the Solution

The inequality \(\frac{2x+6}{x-2}<0\) holds only in the interval \((-3, 2)\), excluding \(x=2\) where the expression is undefined.We include \(-3\) because \(\frac{2(-3)+6}{-3-2} = \frac{0}{-5} = 0\), not less than zero. Thus, \(x=-3\) is not part of the solution and the interval is \((-3, 2)\).

Step 4: Graph the Solution

Graph the solution set on a number line. Represent the interval \((-3, 2)\) with an open circle at \(-3\) and at \(2\), indicating these points are not included in the solution. Shade the space between \(-3\) and \(2\).

Key Concepts

Critical Points

Critical points play a vital role in analyzing and solving nonlinear inequalities. These points are where the expression either becomes zero or undefined. For the inequality \( \frac{2x+6}{x-2}<0 \), we identify the critical points by setting the numerator and denominator separately to zero.

• Numerator: \(2x+6 = 0\) leads to \(x = -3\).
• Denominator: \(x-2 = 0\) results in \(x = 2\).

These values, \(-3\) and \(2\), are critical points because they determine the intervals to test for the inequality. While \(x = -3\) makes the expression zero, \(x = 2\) causes the denominator to become zero, hence rendering the expression undefined. Understanding critical points helps to divide the number line into specific segments to analyze separately.

Interval Notation

Interval notation is a concise way to express a range of solutions for inequalities. After determining the critical points, these points help divide the number line into intervals. For the exercise at hand, the critical points \(-3\) and \(2\) create the intervals: \((-\ infinity, -3)\), \((-3, 2)\), and \((2, \infty)\).

• Use parentheses \((\) and \())\) to indicate that endpoints are not included.
• Use brackets \([\) and \()]\) to indicate that endpoints are included.

In this case, since \( \frac{2x+6}{x-2}<0 \) only holds in \((-3, 2)\), the endpoints \(-3\) and \(2\) are not included, leading to the interval \((-3, 2)\).

Understanding interval notation is key to accurately conveying which parts of the number line satisfy the inequality.

Solution Set

The solution set is the collection of all values that satisfy the given inequality. In solving \( \frac{2x+6}{x-2}<0 \), we use intervals from the critical points \(-3\) and \(2\) to test for where the expression is less than zero.

• For \((-\ infinity, -3)\), a test value like \(-4\) makes the expression positive.
• For \((-3, 2)\), a test value like \(0\) makes the expression negative, satisfying the inequality.
• For \((2, \infty)\), a test value like \(3\) again makes the expression positive.

Only \((-3, 2)\) satisfies the inequality, thus forming our solution set.
Indicating this graphically involves shading the interval between \(-3\) and \(2\) on a number line, excluding the endpoints. The solution set gives us a clear understanding of the range in which the inequality holds true.