Problem 81 - Algebra for College Students

Question

Solve each inequality. State the solution set using interval notation when possible. \(0.23 x^{2}+6.5 x+4.3<0\)

Step-by-Step Solution

Answer

The solution set in interval notation is \(( -28.85, -0.67)\).

Step 1: Recognize the type of inequality

Identify that the given inequality is a quadratic inequality of the form: \(0.23x^{2} + 6.5x + 4.3 < 0\), where the coefficients are: \(a = 0.23\), \(b = 6.5\), and \(c = 4.3\).

Step 2: Calculate the discriminant

The discriminant of a quadratic equation \(ax^{2} + bx + c = 0\) is found using: \( \text{discriminant} = b^2 - 4ac \). Substitute the given values: \(b = 6.5\), \(a = 0.23\), and \(c = 4.3\). Calculate: \( 6.5^2 - 4(0.23)(4.3) = 42.25 - 3.956 = 38.294\).

Step 3: Find the roots using the quadratic formula

The roots of the quadratic equation are found using the formula: \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \).Substitute the values: \( a = 0.23\), \( b = 6.5\), Calculate: \( x = \frac{-6.5 \, \pm \, \sqrt{38.294}}{(2)(0.23)} = \frac{-6.5 \, \pm \, 6.19}{0.46}\)Solve the roots: \( x_1 = \frac{-6.5 + 6.19}{0.46} = -0.67\) and \( x_2 = \frac{-6.5 - 6.19}{0.46} = -28.85\)

Step 4: Test intervals around the roots

The roots \(x_1 = -0.67\) and \(x_2 = -28.85\) divide the number line into three intervals: \(( -\frac{\text{infinity}}{0.23}, -28.85) \), \(( -28.85, -0.67)\), and \((-0.67, \frac{\text{infinity}}{0.23})\).Choose a test point from each interval and substitute it into the inequality \(0.23x^{2} + 6.5x + 4.3 < 0\), to determine where the inequality is true.

Step 5: Determine the solution set

Upon testing, it is found that the inequality is true in the interval between the roots. Thus, the solution set for \frac{ 0.23x^{2} + 6.5x + 4.3 < 0}{0.23} \ in interval notation is: \(( -28.85, -0.67)\)

Key Concepts

Discriminant

The discriminant is a key element in solving quadratic equations. It is derived from the standard form of a quadratic equation, which is: \(ax^{2} + bx + c = 0\). The formula for the discriminant is: \[\text{discriminant} = b^2 - 4ac\].
In our example, the quadratic inequality is \(0.23x^{2} + 6.5x + 4.3 < 0\). Here, \(a = 0.23\), \(b = 6.5\), and \(c = 4.3\).
When we plug these values into the discriminant formula, we get: \[6.5^2 - 4(0.23)(4.3) = 42.25 - 3.956 = 38.294\].
Why is the discriminant important?

• Helps determine the nature of the roots (real or complex).
• Informs us whether a quadratic equation has two distinct roots, one repeated root, or no real roots at all.

In this case, since our discriminant is positive (\(38.294\), it shows us that there are two distinct real roots.

Quadratic Formula

The quadratic formula allows us to find the roots of any quadratic equation. It is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
Using our example \(0.23x^{2} + 6.5x + 4.3 < 0\), we already know that: \(a = 0.23\), \(b = 6.5\), and \(c = 4.3\). Substituting these values into the quadratic formula, we get:
\[ x = \frac{-6.5 \pm \sqrt{38.294}}{2 \cdot 0.23} \ = \frac{-6.5 \pm 6.19}{0.46} \.\].
Solving this, we find the roots to be:

• \(x_1 = \frac{-6.5 + 6.19}{0.46} = -0.67\)
• \(x_2 = \frac{-6.5 - 6.19}{0.46} = -28.85\)

These roots are essential because they help us determine the intervals to test for our inequality.

Interval Notation

Interval notation is a way of representing subsets of the real number line. It is particularly useful when expressing the solution set of inequalities.
We use different types of brackets to indicate whether endpoints are included or not:

• \( (a, b) \): a and b are not included.
• \( [a, b] \): both a and b are included.
• \( (a, b] \): a is not included, but b is.
• \( [a, b) \): a is included, but b is not.

In our example \(0.23x^{2} + 6.5x + 4.3 < 0\), we found the roots to be \(x_1 = -0.67\) and \(x_2 = -28.85\).
These roots divide the number line into three intervals:

• \( (- infinity, -28.85) \)
• \( (-28.85, -0.67) \)
• \( (-0.67, infinity) \)

Upon testing these intervals, we determined that the quadratic inequality is true in the interval between the roots. Thus, the solution set in interval notation is:
\( (-28.85, -0.67) \). This notation is concise and clearly represents the range of x-values that satisfy the inequality.