Problem 9 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ x(2 x+3) \geq 5 $$

Step-by-Step Solution

Answer

The solution to the inequality is \((-\infty, -2.5] \cup [1, \infty)\).

Step 1: Expand the Inequality

Start by expanding the left-hand side of the inequality. The expression is \( x(2x + 3) \). Distribute \( x \) across \( 2x + 3 \) to get:\[ x(2x + 3) = 2x^2 + 3x \]So the inequality becomes:\[ 2x^2 + 3x \, \geq \, 5 \]

Step 2: Move All Terms to One Side

Subtract 5 from both sides of the inequality to set it to zero on one side. This gives:\[ 2x^2 + 3x - 5 \, \geq \, 0 \]

Step 3: Solve the Quadratic Inequality

To solve the inequality, first find the roots of the quadratic equation \( 2x^2 + 3x - 5 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 3 \), and \( c = -5 \).Calculate the discriminant:\[ b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49 \]Calculate the roots:\[ x = \frac{-3 \pm \sqrt{49}}{4} \]\[ x = \frac{-3 \pm 7}{4} \]This gives the roots:\[ x_1 = \frac{4}{4} = 1 \]\[ x_2 = \frac{-10}{4} = -2.5 \]

Step 4: Test Intervals

The roots \( x = 1 \) and \( x = -2.5 \) divide the number line into three intervals: \( (-\infty, -2.5) \), \( (-2.5, 1) \), and \( (1, \infty) \). Test each interval to see where the inequality \( 2x^2 + 3x - 5 \geq 0 \) holds:- For \( x = -3 \) in \( (-\infty, -2.5) \): \[ 2(-3)^2 + 3(-3) - 5 = 18 - 9 - 5 = 4 \] (positive)- For \( x = 0 \) in \( (-2.5, 1) \): \[ 2(0)^2 + 3(0) - 5 = -5 \] (negative)- For \( x = 2 \) in \( (1, \infty) \): \[ 2(2)^2 + 3(2) - 5 = 8 + 6 - 5 = 9 \] (positive)

Step 5: Determine Solution Set

From Step 4, the quadratic expression is non-negative in the intervals \((-\infty, -2.5]\) and \([1, \infty)\). Include the endpoints \(-2.5\) and \(1\) because the inequality is \(\geq\) rather than \(>\). Thus, the solution set in interval notation is:\[ (-\infty, -2.5] \cup [1, \infty) \]

Key Concepts

Quadratic Inequalities

Quadratic inequalities come into play when we deal with expressions of the form \( ax^2 + bx + c \). Solving these inequalities involves finding the values of \( x \) such that the quadratic expression either remains greater than, less than, or equal to a certain value.
A typical problem might require a solution where the expression is greater or equal to zero. Here's the general approach:

• Convert the inequality into a quadratic equation by setting it equal to zero.
• Use methods like factorization or the quadratic formula to find the roots of the equation.
• These roots then define intervals on the number line where you test the inequality.

This helps establish regions where the inequality holds true, thereby solving the inequality.

Discriminant

The discriminant plays a crucial role in understanding the nature of the roots of a quadratic expression. It's denoted as \( b^2 - 4ac \) in the quadratic formula. The value of the discriminant tells us several things:

• If the discriminant is positive, the quadratic has two distinct real roots.
• If it is zero, both roots are real and identical, meaning the parabola touches the x-axis at one point.
• If negative, the quadratic has no real roots, indicating the parabola does not intersect the x-axis.

In the discussed exercise, the discriminant is 49 (positive), confirming two different real roots.

Quadratic Formula

The quadratic formula is a tried and true method for finding the roots of any quadratic equation formulated as \( ax^2 + bx + c = 0 \). The formula looks like this: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
It's incredibly valuable, especially:

• When quadratic equations cannot be factored easily.
• If you need to precisely determine the roots, regardless if they are rational or irrational.

Utilizing this formula requires computing the discriminant first and then substituting the values of \( a \), \( b \), and \( c \). In our scenario, solving \( 2x^2 + 3x - 5 = 0 \) led to the roots \( x = 1 \) and \( x = -2.5 \). These roots serve as critical points for determining the intervals in solving the inequality.

Interval Notation

Once the critical points are identified by solving the quadratic equation, it's necessary to express the solution of the inequality using interval notation. This method efficiently communicates ranges of numbers and is especially useful for inequalities, where not one particular number, but a range, satisfies the equation.
Here's what you need to know:

• Parentheses \( ( ) \) indicate that an endpoint is not included in the solution.
• Brackets \( [ ] \) indicate that an endpoint is included, signified by '>=' or '<=' in the inequality.

In the example, the solution intervals \((-\infty, -2.5] \cup [1, \infty)\) depict that the quadratic inequality holds for values of \( x \) in these specific ranges.