Problem 7 - Algebra A Combined Function

Question

Write the solution set in interval notation. $$ 3 x^{2}+16 x<-5 $$

Step-by-Step Solution

Answer

The solution set is \((-5, -\frac{1}{3})\).

Step 1: Move All Terms to One Side

Start by moving all terms to one side of the inequality to set it to zero. This results in:\[ 3x^2 + 16x + 5 < 0 \]

Step 2: Find the Roots of the Quadratic

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots of the equation \(3x^2 + 16x + 5 = 0\). Here, \(a = 3\), \(b = 16\), and \(c = 5\).1. Calculate the discriminant: \(b^2 - 4ac = 16^2 - 4 \cdot 3 \cdot 5 = 256 - 60 = 196\).2. The roots are: \(x = \frac{-16 \pm \sqrt{196}}{6} = \frac{-16 \pm 14}{6}\).3. Solve for \(x\): * \(x_1 = \frac{-16 + 14}{6} = \frac{-2}{6} = -\frac{1}{3}\) * \(x_2 = \frac{-16 - 14}{6} = \frac{-30}{6} = -5\)

Step 3: Determine Intervals Based on Roots

The roots \(x = -\frac{1}{3}\) and \(x = -5\) divide the x-axis into three intervals: 1. \((-\infty, -5)\)2. \((-5, -\frac{1}{3})\)3. \((-\frac{1}{3}, \infty)\)

Step 4: Test Each Interval

Choose a test point from each interval to determine where the inequality \(3x^2 + 16x + 5 < 0\) holds true.1. For \((-\infty, -5)\), use \(x = -6\): \(3(-6)^2 + 16(-6) + 5 = 108 - 96 + 5 = 17\), which is greater than 0.2. For \((-5, -\frac{1}{3})\), use \(x = -1\): \(3(-1)^2 + 16(-1) + 5 = 3 - 16 + 5 = -8\), which is less than 0.3. For \((-\frac{1}{3}, \infty)\), use \(x = 0\): \(3(0)^2 + 16(0) + 5 = 5\), which is greater than 0.

Step 5: Write the Solution in Interval Notation

Since the quadratic expression is less than zero in the interval \((-5, -\frac{1}{3})\), the solution set in interval notation is:\((-5, -\frac{1}{3})\)

Key Concepts

Quadratic Inequalities

Quadratic inequalities are mathematical expressions involving a quadratic equation accompanied by inequality symbols like \(<\), \(>\), \(\leq\), or \(\geq\). While a quadratic equation is in the form \(ax^2 + bx + c = 0\), a quadratic inequality would be \(ax^2 + bx + c < 0\) or similar.To solve a quadratic inequality, follow these essential steps:

• Move all terms to one side to set the inequality to zero. This makes it easier to analyze.
• Find the roots of the equivalent quadratic equation \(ax^2 + bx + c = 0\) to divide the x-axis into intervals.
• Test each interval to check where the inequality holds true.

By following these steps, you can effectively determine the intervals where the inequality expression meets the given criteria.

Interval Notation

Interval notation is a way to express the set of solutions for equations and inequalities concisely. By using special symbols, we can clearly denote the range of values, helping to communicate the solutions quickly.In interval notation:

• Parentheses \(()\) indicate that an endpoint is not included, representing an open interval.
• Square brackets \([]\) signify that an endpoint is included, marking a closed interval.
• "Union" symbol \(\cup\) combines disjoint intervals into a single set.

For the quadratic inequality example provided, the interval notation result was \((-5, -\frac{1}{3})\). This implies the solution set includes all numbers between -5 and -\(\frac{1}{3}\) but excludes the endpoints themselves.

Quadratic Formula

The quadratic formula is a pivotal tool for solving quadratic equations. It provides an exact solution to any quadratic equation of the form \(ax^2 + bx + c = 0\). The quadratic formula is defined as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Breaking it down:

• \(b^2 - 4ac\) is the discriminant, determining the nature of the roots.
• A positive discriminant results in two distinct real roots.
• A zero discriminant yields one real root (a repeated root).
• A negative discriminant produces complex roots.

For our exercise, the discriminant was positive (196), giving us two real roots: \(-\frac{1}{3}\) and \(-5\), which were crucial for identifying the valid interval in solving the inequality.

Solution Set

The solution set of a quadratic inequality encompasses all the values of \(x\) that satisfy the inequality. Once the intervals have been tested for truth, formulating the solution set allows us to express these values succinctly.To derive the solution set:

• Determine where the quadratic expression is negative (or positive, depending on the inequality).
• Use test points to check the sign of the expression in each interval.
• Combine all intervals where the expression fulfills the inequality condition.

In the given example, the solution set for the inequality \(3x^2 + 16x + 5 < 0\) was found to be \((-5, -\frac{1}{3})\). Hence, only values of \(x\) within this interval meet the condition of the inequality.