Problem 37 - College Algebra with Modeling and Visualization

Question

Solve the inequality. $$ 6 x^{2}-x<1 $$

Step-by-Step Solution

Answer

Solution: \(-\frac{1}{3} < x < \frac{1}{2}\).

Step 1: Rewrite the Inequality

Start by moving all terms to one side of the inequality to form a standard quadratic inequality. This results in: \[ 6x^2 - x - 1 < 0 \] which is in the form \( ax^2 + bx + c < 0 \).

Step 2: Factor the Quadratic Expression

Next, factor the quadratic expression \( 6x^2 - x - 1 \). Set up the factors as \( (ax + b)(cx + d) \). Through trial and method or using the quadratic formula, we find: \[ 6x^2 - x - 1 = (3x + 1)(2x - 1) \].

Step 3: Determine the Roots of the Quadratic Expression

To find the roots (or critical points) of the inequality, set each factor equal to zero: \( 3x + 1 = 0 \) which gives \( x = -\frac{1}{3} \), and \( 2x - 1 = 0 \) which gives \( x = \frac{1}{2} \).

Step 4: Test Intervals Around the Roots

The roots divide the number line into intervals: \( (-\infty, -\frac{1}{3}) \), \( (-\frac{1}{3}, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \). Choose a test point from each interval to determine where the inequality \( (3x + 1)(2x - 1) < 0 \) is true.- For \((-\infty, -\frac{1}{3})\), test point \(x = -1\): \((3(-1) + 1)(2(-1) - 1) = (-2)(-3) = 6\) (not less than 0)- For \((-\frac{1}{3}, \frac{1}{2})\), test point \(x = 0\): \((3(0) + 1)(2(0) - 1) = 1(-1) = -1\) (less than 0)- For \((\frac{1}{2}, \infty)\), test point \(x = 1\): \((3(1) + 1)(2(1) - 1) = (4)(1) = 4\) (not less than 0)

Step 5: State the Solution Set

From the test intervals, the solution is in the interval \((-\frac{1}{3}, \frac{1}{2})\) where the inequality holds true. Therefore, the solution set is: \(-\frac{1}{3} < x < \frac{1}{2}\).

Key Concepts

Factoring Quadratics

Factoring quadratics is a method used to simplify expressions like \( ax^2 + bx + c \) into a product of two linear terms. This is incredibly useful when trying to solve quadratic equations or inequalities. In the exercise, we started with the expression \( 6x^2 - x - 1 \) and aimed to rewrite it as a product of factors. This helps because it reveals the values of \( x \) that make the expression equal to zero, known as roots or critical points.

There are several ways to factor quadratic expressions:

• Trial and Error: This involves guessing pairs of numbers that multiply to give the ac term (in this case, \( 6 imes -1 = -6 \)) and add to give the b term (here, -1).
• Quadratic Formula: Though not direct factoring, using it can guide us in factoring. It's given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), allowing us to find roots that work as guides for factoring.
• Decomposition: Split the middle term into two terms that allow the expression to be grouped and factored by grouping.

In the example, we discover that \( 6x^2 - x - 1 = (3x + 1)(2x - 1) \). Successfully factoring this way simplifies further solving immensely.

Critical Points

Critical points in a quadratic inequality are the x-values where the expression changes from positive to negative or vice versa. These are found by setting the factored expression to zero. For the factored inequality \((3x + 1)(2x - 1) < 0\), the roots indicate when each individual factor equals zero.

By solving the equations \( 3x + 1 = 0 \) and \( 2x - 1 = 0 \), we find our critical points:

• \( x = -\frac{1}{3} \)
• \( x = \frac{1}{2} \)

These roots tell us that at \( x = -\frac{1}{3} \) and \( x = \frac{1}{2} \), the expression \((3x + 1)(2x - 1)\) equals zero. Hence, the behavior of the inequality can change at these points.

It's crucial to plot or note these critical points when determining where the inequality holds true. They offer a roadmap of where to test intervals on the number line to see where the original inequality is satisfied.

Interval Testing

Once we have identified the critical points, interval testing helps us determine which intervals between these points satisfy the inequality. The number line is divided into intervals by these critical points. In our example, these intervals are:

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

To decide if the inequality \((3x + 1)(2x - 1) < 0\) holds in a specific interval, we pick a test point from each section and substitute it back into the inequality.

Here's what was found:

• Testing \(x = -1\) in \((-\infty, -\frac{1}{3})\), the product is greater than 0.
• Testing \(x = 0\) in \((-\frac{1}{3}, \frac{1}{2})\), the product is less than 0.
• Testing \(x = 1\) in \((\frac{1}{2}, \infty)\), the product is greater than 0.

Only in the interval \((-\frac{1}{3}, \frac{1}{2})\) is the inequality \(less\) than zero, indicating it's where our solution set lies. The solution is then \(-\frac{1}{3} < x < \frac{1}{2}\), providing clarity on which values of \(x\) satisfy our quadratic inequality.