Problem 28 - Algebra for College Students

Question

Solve each inequality. $$12 x^{2}-20 x+3 \geq 0$$

Step-by-Step Solution

Answer

The solution is \( x \in (-\infty, \frac{1}{6}] \cup [\frac{3}{2}, \infty) \).

Step 1: Identify the type of inequality

This inequality is a quadratic inequality in the form of \( ax^2 + bx + c \geq 0 \). We have \( a = 12 \), \( b = -20 \), and \( c = 3 \).

Step 2: Find the roots of the quadratic equation

To solve \( 12x^2 - 20x + 3 = 0 \), we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting the values, we get:\[ x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot 12 \cdot 3}}{2 \cdot 12} \]Simplifying further:\[ x = \frac{20 \pm \sqrt{400 - 144}}{24} \]\[ x = \frac{20 \pm \sqrt{256}}{24} \]\[ x = \frac{20 \pm 16}{24} \]Thus, the roots are \( x = \frac{36}{24} = \frac{3}{2} \) and \( x = \frac{4}{24} = \frac{1}{6} \).

Step 3: Determine the intervals for testing

The roots \( x = \frac{3}{2} \) and \( x = \frac{1}{6} \) divide the number line into three intervals: \(( -\infty, \frac{1}{6} )\), \(( \frac{1}{6}, \frac{3}{2} )\), and \(( \frac{3}{2}, \infty )\).

Step 4: Test each interval

Choose a test point from each interval and substitute it into the inequality:- For \( x = 0 \) in \( (-\infty, \frac{1}{6}) \): \( 12(0)^2 - 20(0) + 3 = 3 \geq 0 \) - True- For \( x = 1 \) in \( (\frac{1}{6}, \frac{3}{2}) \): \( 12(1)^2 - 20(1) + 3 = -5 \geq 0 \) - False- For \( x = 2 \) in \( (\frac{3}{2}, \infty) \): \( 12(2)^2 - 20(2) + 3 = 19 \geq 0 \) - True

Step 5: Combine the solutions

Since the inequality holds for the intervals \((-\infty, \frac{1}{6}]\) and \([\frac{3}{2}, \infty)\), and both boundary points satisfy \( 12x^2 - 20x + 3 = 0 \), they are included in the solution set. Thus, the solution is: \( x \in (-\infty, \frac{1}{6}] \cup [\frac{3}{2}, \infty) \).

Key Concepts

Solving Quadratic Equations

Quadratic equations form the basis of understanding quadratic inequalities. A quadratic equation looks like this: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable we are solving for. Solving quadratic equations involves finding values of \( x \) that satisfy this equation. These values are often called 'roots' or 'solutions'.

To solve a quadratic equation, we can use various methods, but the quadratic formula is one of the most reliable methods since it applies to all types of quadratics. It doesn't require factorization or special tricks, making it widely applicable across different problems.

Once we find the roots, these values play a crucial role in determining the number line's 'critical points'. These are essential for solving quadratic inequalities, as they divide the number line into sections to test whether the inequality holds true.

Interval Testing

Interval testing is a strategy used once we have the roots of a quadratic equation. It involves dividing the number line into intervals and testing each section to see whether it satisfies the inequality. For instance, if our roots are \( x = \frac{3}{2} \) and \( x = \frac{1}{6} \), the number line is split into these intervals:

• \((-\infty, \frac{1}{6})\)
• \((\frac{1}{6}, \frac{3}{2})\)
• \((\frac{3}{2}, \infty)\)

By choosing test points from each interval and substituting them back into the original inequality, we determine whether the inequality holds in those ranges.

The sign of the quadratic expression often alternates between these intervals. For example, if a point within an interval makes the inequality true, then all points in that interval likely do. This process helps us piece together where the overall solution resides along the number line. Remember, our aim is to include only those intervals where the inequality satisfies the conditions stated. If the inequality is non-inclusive (e.g., \( > \) or \( < \)), endpoints of intervals are not included in the solution set.

Quadratic Formula

The quadratic formula is a powerful tool allowing us to find the roots of any quadratic equation. It is written as: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula systematically uses the coefficients \( a \), \( b \), and \( c \) from the quadratic equation \( ax^2 + bx + c = 0 \).

To break it down:

• The term \(-b\) changes the sign of the \( b \) coefficient.
• \( \sqrt{b^2 - 4ac} \) calculates the 'discriminant'. It tells you key details about the nature of the roots:
• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is one real root (a repeated root).
• If \( b^2 - 4ac < 0 \), the roots are complex (not real).
• Finally, the expression is divided by \( 2a \), as all parts of the quadratic equation relate to the \( a \) coefficient.

When used in solving inequalities, finding these roots is just the beginning. They help pinpoint the crucial points needed for further analysis.