Problem 5 - Algebra A Combined Function

Question

Write the solution set in interval notation. $$ x^{2}+8 x+15 \geq 0 $$

Step-by-Step Solution

Answer

The solution set is \((- fty, -5] \cup [-3, fty)\).

Step 1: Solve the Corresponding Equation

To find the critical points, solve the equation \( x^2 + 8x + 15 = 0 \). This is a quadratic equation that can be factored as \( (x + 3)(x + 5) = 0 \). Setting each factor equal to zero gives us the roots \( x = -3 \) and \( x = -5 \).

Step 2: Determine the Number Line Intervals

The roots divide the number line into three intervals: \((-fty, -5)\), \((-5, -3)\), and \((-3, fty)\). We will test each interval to determine where the inequality \(x^2 + 8x + 15 \geq 0\) holds.

Step 3: Test Each Interval

Choose a test point from each interval:1. For \((-fty, -5)\), use \(x = -6\). Calculate \((x + 3)(x + 5) = (-6 + 3)(-6 + 5) = (-3)(-1) = 3\), which is positive.2. For \((-5, -3)\), use \(x = -4\). Calculate \((-4 + 3)(-4 + 5) = (-1)(1) = -1\), which is negative.3. For \((-3, fty)\), use \(x = 0\). Calculate \((0 + 3)(0 + 5) = 3\cdot5 = 15\), which is positive.

Step 4: Evaluate Boundary Points

Check the boundary points \(x = -5\) and \(x = -3\). Since the inequality is \( \geq 0 \), both boundary points, where the expression equals zero, are included in the solution set.

Step 5: Write the Solution in Interval Notation

Based on the previous steps, the solution set in interval notation is \((-fty, -5] \cup [-3, fty)\), including both \(-5\) and \(-3\) since the inequality is non-strict.

Key Concepts

Factoring Quadratics

Factoring quadratics is a crucial skill for solving quadratic inequalities. It involves rewriting a quadratic expression in the form \( ax^2 + bx + c \) as a product of two linear factors. For the inequality \( x^2 + 8x + 15 \geq 0 \), the quadratic can be factored into \((x + 3)(x + 5) \). This factorization reveals the roots or solutions of the equation when the product equals zero.To factor a simple quadratic like this one, you look for two numbers that multiply to the constant term (here, 15) and add up to the coefficient of \(x\) (which is 8). In this case, those numbers are 3 and 5. Thus, the quadratic \( x^2 + 8x + 15 \) can be expressed as \( (x + 3)(x + 5) \). This helps in determining critical points for solving inequalities.

Critical Points

Critical points are the values of \(x\) that satisfy the equation when the product of factors equals zero. They are important because they divide the number line into segments that you can test.For the quadratic \( x^2 + 8x + 15 = 0 \), the factors \((x + 3)(x + 5) = 0\) provide the critical points \( x = -3 \) and \( x = -5 \). These points are critical because they mark where the sign of the expression might change.By solving \( x^2 + 8x + 15 = 0 \), we found the critical points \( -3 \) and \( -5 \). These are used to determine intervals on the number line that need testing to identify where the quadratic inequality holds true or becomes zero. Each segment gives different potential outcomes for the inequality.

Interval Notation

Interval notation is a concise way to express sets of numbers that satisfy inequalities. It uses brackets and parentheses to describe intervals on the number line.In the context of the inequality \( x^2 + 8x + 15 \geq 0 \), after testing each interval divided by the critical points \(-3\) and \(-5\), we found that the solution is \((-\infty, -5] \cup [-3, \infty)\). This notation indicates:

• \((-\infty, -5]\) means starting from negative infinity up to and including \(-5\).
• \([-3, \infty)\) means starting from \(-3\), including it, and extending to positive infinity.

Understanding interval notation helps in cleanly expressing solutions to inequalities. It clearly designates which numbers satisfy the given inequality, whether they include boundaries (with square brackets) or exclude them (with parentheses).