Problem 27 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ 4 x^{2}+9 x<9 $$

Step-by-Step Solution

Answer

The solution set is \( (-3, \frac{3}{4}) \).

Step 1 - Rewrite the Inequality

Start by rewriting the inequality in the standard form: \[ 4x^2 + 9x - 9 < 0 \]

Step 2 - Find the Roots of the Quadratic Equation

To determine where the inequality changes sign, solve the corresponding equation: \[ 4x^2 + 9x - 9 = 0 \] Use the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 4 \), \( b = 9 \), and \( c = -9 \).

Step 3 - Calculate the Discriminant

Calculate the discriminant \( \Delta \): \[ \Delta = b^2 - 4ac \] \[ \Delta = 9^2 - 4(4)(-9) = 81 + 144 = 225 \]

Step 4 - Find the Roots

Substitute \( \Delta = 225 \) back into the quadratic formula: \[ x = \frac{-9 \pm \sqrt{225}}{8} \] \[ x = \frac{-9 \pm 15}{8} \] So the roots are: \[ x_1 = \frac{6}{8} = \frac{3}{4} \] \[ x_2 = \frac{-24}{8} = -3 \]

Step 5 - Determine the Intervals for Testing

The inequality \( 4x^2 + 9x - 9 < 0 \) changes sign at the roots. Check the inequality in the intervals: \[ (-\infty, -3), (-3, \frac{3}{4}), (\frac{3}{4}, \infty) \]

Step 6 - Test Each Interval

Select a test point in each interval and determine if it satisfies the inequality. For example:1. For \( x = -4 \) in \((-\infty, -3)\): \[ 4(-4)^2 + 9(-4) - 9 = 64 - 36 - 9 = 19 > 0 \] Thus, \((-\infty, -3)\) is not part of the solution.2. For \( x = 0 \) in \((-3, \frac{3}{4})\): \[ 4(0)^2 + 9(0) - 9 = -9 < 0 \] Thus, \((-3, \frac{3}{4})\) is part of the solution.3. For \( x = 1 \) in \((\frac{3}{4}, \infty)\): \[ 4(1)^2 + 9(1) - 9 = 4 + 9 - 9 = 4 > 0 \] Thus, \((\frac{3}{4}, \infty)\) is not part of the solution.

Step 7 - Write the Solution Set

Combine the results from testing the intervals. The inequality \( 4x^2 + 9x - 9 < 0 \) holds in the interval \[ (-3, \frac{3}{4}) \].

Step 8 - Illustrate on the Real Number Line

Draw a number line and shade the interval \( (-3, \frac{3}{4}) \). Use open circles at \( -3 \) and \( \frac{3}{4} \) to indicate that these points are not included in the solution.

Key Concepts

Quadratic Formula

To use the formula, identify the coefficients:

• $The\; coefficient$a in front of x^2.
• The coefficient b in front of x.
• The constant term c.

Now, plug values for a, b, and c into the formula to find the values of x.

Discriminant

Calculate the discriminant is a straightforward process. All you need is to plug the values of a, b, and c into the discriminant formula.

Solution Interval

Once you have the roots of the quadratic equation, which are the points where the function equals zero, you can figure out where the entire expression is less than zero (or satisfy the inequality).Let's break down this step by step.First, write down the intervals which are formed by these roots on the number line. For example, if the roots are -3 and \t\textTime\text\t\text\frac{3}{4}\text\t\text\t\text\t\text\t\t\text\tThese intervals would be (-∞, -3), (-3, \frac{3},{4})and (∞, \frac{3}{4},)
Next, select values from these intervals and plug them into the original inequality..Check if the resulting expression is true within that interval.
In our example:

• By selecting &150;
-∞, -3), for x=-4\textThe expression, 4(-4)^2 + 9(-4) – 91) Both yields a positive value .
• Choose a number from (-3, \(\frac{3}{4}\)), for x=0
By plugging this value into the inequality we get,
the result of negative value of -9 <0ii showing that\(\text(\text)\text(\text(\frac{3}{4}\) interval should be the solution.
• Choose a number from (∞, x=4)Plugging in this vale to the inequality again results in positive 4>0. thus, this interval is not part of the solution So, the final range of z< (-3≤3,4)

After finding the intervals that satisfy the inequality, you can bsolve them.
This set can be described using interval-notation.

Interval Notation

Interval notation is a concise way to describe intervals on the number line. It includes all the numbers between two endpoints.
To understand this better, let's break it down:
When we have a solution like (-3, ⚡\(\frac{3}{4}\)), we write it as a shortend form showing that all numbers between -3 and \(\frac{3}{4}\) all belong to the solutions set.
In the interval I notation,
(use open parenthesis whenever an endpoint a< is not including. For closed point such as [including use

In our example, the solution set is (<-3,3/4 Starting with -3 and end & usat the turning point of ¾)Your final solution should be written:
(-3,\(\frac{3}{4}\)) .
Moreover, open circles are used to indicate that endpoints -3 &3,4x are not part of the point set .
This averted roots give us the complete soloption showing that that equation has -3<3/4 has 0 (solution denoting that equation>$ of solution should be used in that context completing the meaning will g example intervals.
:

make sure to correctly represent them