Problem 52 - Beginning and Intermediate Algebra

Question

Solve each inequality, and graph the solution set. $$ \frac{9 x-8}{4 x^{2}+25}<0 $$

Step-by-Step Solution

Answer

The solution set is \( (-\infty, \frac{8}{9}) \) and is represented with a shaded region to the left of \( \frac{8}{9} \) on a number line.

Step 1: Identify Critical Points

Find the values of x that make the numerator or the denominator equal to zero. Solve for x when the numerator is zero: \[ 9x - 8 = 0 \] \[ x = \frac{8}{9} \]. Next, solve for x when the denominator is zero: \[ 4x^2 + 25 = 0 \]. This equation has no real solutions as the sum of squares is always positive.

Step 2: Determine Intervals

The critical point divides the real number line into two intervals: \((-\infty, \frac{8}{9})\) and \((\frac{8}{9}, \infty)\).

Step 3: Test Points in Each Interval

Choose a test point from each interval and substitute back into the inequality: For interval \((-\infty, \frac{8}{9})\), let’s use x = 0:\[ \frac{9(0) - 8}{4(0)^2 + 25} = \frac{-8}{25} < 0 \].For interval \((\frac{8}{9}, \infty)\), let’s use x = 1:\[ \frac{9(1) - 8}{4(1)^2 + 25} = \frac{1}{29} > 0 \].

Step 4: Analyze Critical Points

Since the inequality specifies less than zero and not less than or equal to zero, the solution does not include the critical point \( \frac{8}{9} \).

Step 5: Write the Solution

The solution of the inequality is all x-values in the interval\(-\infty < x < \frac{8}{9} \).

Step 6: Graph the Solution

To graph the solution, draw a number line. Identify the point \( \frac{8}{9} \) on the number line and use an open circle to indicate that this point is not included. Shade the region to the left of \(\frac{8}{9}\) to illustrate the solution set \( (-\infty, \frac{8}{9}) \).

Key Concepts

critical points

Critical points are vital when you're solving inequalities. These points are where the expression may change its sign. For rational inequalities, the critical points occur when either the numerator or the denominator is zero. In our example, the numerator is zero when:
\[ 9x - 8 = 0 \]
Solving this equation for x, we get
\[ x = \frac{8}{9} \].

The denominator of this inequality is
\[ 4x^2 + 25 \].
However, if you try to solve \( 4x^2 + 25 = 0 \), you'll see that there are no real solutions since the sum of a square and a positive number cannot be zero. So, the only critical point here is \( x = \frac{8}{9} \).

intervals

Using the critical points, we can determine the intervals to test. A critical point divides the number line into separate intervals. For our example, \( x = \frac{8}{9} \) splits the real number line into two intervals:
\( (-\infty, \frac{8}{9}) \) and \( (\frac{8}{9}, \infty) \).

These intervals can guide us to see where the expression in the inequality is positive or negative. Next, we choose a test point from each interval and substitute it back into the original inequality to check the sign.

solution set

Once you've tested the intervals, you need to determine which parts of the number line satisfy the inequality. For our example:

Let's consider test points. Choose \( x = 0 \) for the interval \( (-\infty, \frac{8}{9}) \):
\[ \frac{9(0) - 8}{4(0)^2 + 25} = \frac{-8}{25} < 0 \].
This shows that the inequality is true in \( (-\infty, \frac{8}{9}) \).

Now, choose \( x = 1 \) for the interval \( (\frac{8}{9}, \infty) \):
\[ \frac{9(1) - 8}{4(1)^2 + 25} = \frac{1}{29} > 0 \].
This shows that the inequality is false in \( (\frac{8}{9}, \infty) \).

Therefore, our solution set is \( (-\infty, \frac{8}{9}) \). Remember, because the inequality is strictly less than zero, the critical point \( \frac{8}{9} \) is not included. So, we represent it with an open interval.

graphing inequalities

Graphing inequalities helps visualize the solution set. Here is a simple guide to graphing the solution for our example:

* Draw a number line.
* Mark the critical point \( \frac{8}{9} \) on the number line. Use an open circle at \( \frac{8}{9} \) to show that it is not included in the solution set.
* Shade the region to the left of \( \frac{8}{9} \), indicating that all values less than \( \frac{8}{9} \) are part of the solution set.

The shaded section represents \( (-\infty, \frac{8}{9}) \), which is where the original inequality holds true. Using this graphical method makes it easier to understand and communicate the solution effectively.