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Magazine Chapter 11: Problem 33
Solve each inequality. Write the solution set in interval notation. $$ 6 x^{2}-5 x \geq 6 $$
Short Answer
Expert verified
The solution set in interval notation is \((-\infty, -\frac{2}{3}] \cup [\frac{3}{2}, \infty)\).
Step by step solution
01
Rearrange the Inequality
Firstly, rearrange the inequality into a standard form. We move all terms to one side of the inequality to get a quadratic inequality in the form of \( ax^2 + bx + c \), where \( c \) is on the other side. Thus, this becomes: \[ 6x^2 - 5x - 6 \geq 0 \]
02
Factor the Quadratic Expression
Next, we factor the quadratic expression. We are looking for two numbers that multiply to \(-36\) (\(6\) times \(-6\)) and add to \(-5\). These numbers are \(-9\) and \(+4\). Hence, we factor the quadratic as: \[ (2x - 3)(3x + 2) \geq 0 \]
03
Determine the Roots
Find the roots of the equation \((2x - 3)(3x + 2) = 0\). Solving \(2x - 3 = 0\) gives us \(x = \frac{3}{2}\), and Solving \(3x + 2 = 0\) gives us \(x = -\frac{2}{3}\). These roots divide the number line into intervals.
04
Test Each Interval
Test a point from each interval to determine where the inequality \((2x - 3)(3x + 2) \geq 0\) holds true:1. For \(x < -\frac{2}{3}\), pick \(x = -1\). \((2(-1) - 3)(3(-1) + 2) = (-5)(-1) > 0\).2. For \(-\frac{2}{3} < x < \frac{3}{2}\), pick \(x = 0\). \((2(0) - 3)(3(0) + 2) = (-3)(2) < 0\).3. For \(x > \frac{3}{2}\), pick \(x = 2\). \((2(2) - 3)(3(2) + 2) = (1)(8) > 0\).
05
Write the Solution in Interval Notation
From the intervals tested, the inequality is true for \(x < -\frac{2}{3}\) and \(x > \frac{3}{2}\), including where it equals zero. Thus, the solution is: \( x \in \left(-\infty, -\frac{2}{3}\right] \cup \left[\frac{3}{2}, \infty\right) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interval Notation
Interval notation is a way to express a range of numbers along the number line where an inequality holds true. It gives a concise description of solution sets, using parentheses and brackets to indicate whether endpoints are included.
- Parentheses \( \( \) \) signify that an endpoint is not included in the interval. This is used when the inequality is strictly less than or greater than (">" or "<").
- Brackets \( \[ \] \) imply that an endpoint is included. This is used for less than or equal to, or greater than or equal to ("鈮�" or "鈮�").
- For \(x < -\frac{2}{3}\), the interval is \((-\infty, -\frac{2}{3}]\), as \(-\frac{2}{3}\) is included.
- For \(x > \frac{3}{2}\), \([\frac{3}{2}, \infty)\), as \(\frac{3}{2}\) is also included.
Factoring Quadratic Equations
Factoring quadratic equations is an essential step in solving quadratic inequalities. When factoring, the goal is to express the quadratic equation in the form of a product of binomials. This makes it easier to find the roots and solve the inequality.
Here鈥檚 how to approach factoring:
Here鈥檚 how to approach factoring:
- Identify coefficients \(a\), \(b\), and \(c\) from the quadratic equation \(ax^2 + bx + c\).
- Find two numbers that multiply to \(a \cdot c\) and add to \(b\).
- Rewrite the quadratic expression to split the middle term using the two numbers found.
- Factor by grouping to find the binomials.
- Calculate \(6 \times (-6) = -36\).
- Find two numbers: \(-9\) and \(+4\) to split \(-5x\).
- Rewrite: \(6x^2 - 9x + 4x - 6\).
- Group and factor: \((2x-3)(3x+2)\).
Inequality Solution Sets
In solving quadratic inequalities, the solution set is composed of all the values of \(x\) that make the inequality true. Determining this set involves several steps.
Here's a simplified guide to finding inequality solution sets:
Here's a simplified guide to finding inequality solution sets:
- Find roots: Solve the factored equation for roots, which are transition points where the inequality may change. These are \(x = \frac{3}{2}\) and \(x = -\frac{2}{3}\) in our example.
- Divide the number line: Use these roots to divide the number line into separate intervals.
- Test intervals: Select test points from each interval and substitute back into the factored inequality to see if they satisfy it. Confirm whether the inequality holds (\(\geq 0\) or \(\leq 0\)).
- Compile the solution set: Combine intervals where the inequality holds. For instance, for \((2x-3)(3x+2) \geq 0\), the solution intervals are \((-\infty, -\frac{2}{3}]\) and \([\frac{3}{2}, \infty)\).
