Solving Rational Inequalities is an essential skill in high school mathematics, especially in preparation for advanced topics like calculus and college algebra. This page provides a detailed, step-by-step approach to solving rational inequalities, including how to find critical points, analyze sign charts, and determine solution intervals. With clear explanations and worked-out examples, this lesson is designed to help Grade 12 students understand not just the how, but also the why behind each step. Whether you're reviewing for an exam or building a deeper understanding, this page offers the support you need to master rational inequalities.
Step 1: Identify the critical points (zeros of numerator and denominator)
Critical points occur where the expression is zero or undefined.
Numerator: \( x - 2 = 0 \Rightarrow x = 2 \)
Denominator: \( x + 1 = 0 \Rightarrow x = -1 \)
So the critical points are: \[ x = -1 \quad \text{(undefined)}, \quad x = 2 \quad \text{(zero)} \]
Step 2: Divide the number line into intervals
Use the critical points to divide the number line:
Interval 1: \( (-\infty, -1) \)
Interval 2: \( (-1, 2) \)
Interval 3: \( (2, \infty) \)
Step 3: Test a point in each interval
Choose a value in each interval and test the sign of the expression.
Interval 1: \( x = -2 \) \[ \dfrac{-2 - 2}{-2 + 1} = \dfrac{-4}{-1} = 4 > 0 \]
Interval 2: \( x = 0 \) \[ \dfrac{0 - 2}{0 + 1} = \dfrac{-2}{1} = -2 \lt 0 \]
Interval 3: \( x = 3 \) \[ \dfrac{3 - 2}{3 + 1} = \dfrac{1}{4} > 0 \]
Step 4: Include or exclude the critical points
\( x = -1 \) makes the denominator zero and is therefore excluded
\( x = 2 \) makes the numerator zero which satisfies \( \ge 0 \) and is therefore included.
Finally, The solution set of the given ineqiuality is: \[ (-\infty, -1) \cup [2, \infty) \]
Step 1: Rewrite the inequality with zero on one side \[ \dfrac{x + 1}{x + 3} - 2 \le 0 \]
Rewrite with a common denominator: \[ \dfrac{x + 1 - 2(x + 3)}{x + 3} \le 0 \]
Simplify the numerator: \[ \dfrac{x + 1 - 2x - 6}{x + 3} = \dfrac{-x - 5}{x + 3} \]
So the inequality becomes: \[ \dfrac{-x - 5}{x + 3} \le 0 \]
Step 2: Identify the critical points (zeros of numerator and denominator)
Numerator: \( -x - 5 = 0 \Rightarrow x = -5 \)
Denominator: \( x + 3 = 0 \Rightarrow x = -3 \)
So the critical points are: \[ x = -5 \quad \text{(zero)}, \quad x = -3 \quad \text{(undefined)} \]
Step 3: Divide the number line into intervals
Use the critical points to divide the number line:
Interval 1: \( (-\infty, -5) \)
Interval 2: \( (-5, -3) \)
Interval 3: \( (-3, \infty) \)
Step 4: Test a point in each interval
Choose a value in each interval and test the sign of the expression.
Interval 1: \( x = -6 \) \[ \dfrac{-(-6) - 5}{-6 + 3} = \dfrac{6 - 5}{-3} = \dfrac{1}{-3} = -\dfrac{1}{3} \lt 0 \]
Interval 2: \( x = -4 \) \[ \dfrac{-(-4) - 5}{-4 + 3} = \dfrac{4 - 5}{-1} = \dfrac{-1}{-1} = 1 \gt 0 \]
Interval 3: \( x = 0 \) \[ \dfrac{-0 - 5}{0 + 3} = \dfrac{-5}{3} \lt 0 \]
Step 5: Include or exclude the critical points
\( x = -3 \) makes the denominator zero and is therefore excluded.
\( x = -5 \) makes the numerator zero which satisfies \( \le 0 \) and is therefore included.
Finally, the solution set of the given inequality is: \[ (-\infty, -5] \cup (-3, \infty) \]
Step 1: Factor numerator and denominator \[ 4x^2 + 5x - 9 = (4x + 9)(x - 1) \] \[ x^2 - x - 6 = (x - 3)(x + 2) \]
So the inequality becomes: \[ \dfrac{(4x + 9)(x - 1)}{(x - 3)(x + 2)} \ge 0 \]
Step 2: Identify the critical points (zeros of numerator and denominator)
Numerator:
Denominator:
So the critical points are: \[ x = -\dfrac{9}{4}, \quad x = -2, \quad x = 1, \quad x = 3 \]
Step 3: Arrange the critical points on the number line
From left to right on the number line: \[ x = -\infty, \quad -\dfrac{9}{4}, \quad -2, \quad 1, \quad 3, \quad +\infty \]
So the intervals are:
Step 4: Test a point in each interval
Interval 1: \( x = -3 \) \[ \dfrac{(4(-3) + 9)(-3 - 1)}{(-3 - 3)(-3 + 2)} = \dfrac{(-12 + 9)(-4)}{(-6)(-1)} = \dfrac{(-3)(-4)}{6} = \dfrac{12}{6} = 2 > 0 \]
Interval 2: \( x = -2.1 \) \[ \dfrac{(4(-2.1) + 9)(-2.1 - 1)}{(-2.1 - 3)(-2.1 + 2)} \approx -3.64705\dots \lt 0 \]
Interval 3: \( x = 0 \) \[ \dfrac{(4(0) + 9)(0 - 1)}{(0 - 3)(0 + 2)} = \dfrac{(9)(-1)}{(-3)(2)} = \dfrac{-9}{-6} = 1.5 > 0 \]
Interval 4: \( x = 2 \) \[ \dfrac{(4(2) + 9)(2 - 1)}{(2 - 3)(2 + 2)} = \dfrac{(8 + 9)(1)}{(-1)(4)} = \dfrac{17}{-4} = -4.25 \lt 0 \]
Interval 5: \( x = 4 \) \[ \dfrac{(4(4) + 9)(4 - 1)}{(4 - 3)(4 + 2)} = \dfrac{(16 + 9)(3)}{(1)(6)} = \dfrac{75}{6} > 0 \]
Step 5: Include or exclude the critical points
Step 6: Determine where the expression is ≥ 0
From testing and point inclusion:
Final Answer: \[ (-\infty, -\dfrac{9}{4}] \cup (-2, 1] \cup (3, \infty) \]
Step 1: Factor the expression
Numerator: \( x^3 + 1 = (x + 1)(x^2 - x + 1) \)
Denominator: \( x^2 - 9 = (x - 3)(x + 3) \)
So the inequality becomes: \[ \dfrac{(x + 1)(x^2 - x + 1)}{(x - 3)(x + 3)} < 0 \]
Step 2: Identify the critical points (zeros of numerator and denominator)
\( x + 1 = 0 \Rightarrow x = -1 \)
\( x - 3 = 0 \Rightarrow x = 3 \)
\( x + 3 = 0 \Rightarrow x = -3 \)
Note: \( x^2 - x + 1 \) has no real roots because its discriminant is negative: \[ \Delta = (-1)^2 - 4(1)(1) = -3 \]
So the critical points are: \[ x = -3 \quad \text{(undefined)}, \quad x = -1 \quad \text{(zero)}, \quad x = 3 \quad \text{(undefined)} \]
Step 3: Divide the number line into intervals
Use the critical points to divide the number line:
Interval 1: \( (-\infty, -3) \)
Interval 2: \( (-3, -1) \)
Interval 3: \( (-1, 3) \)
Interval 4: \( (3, \infty) \)
Step 4: Test a point in each interval
Interval 1: \( x = -4 \) \[ \dfrac{(x + 1)(x^2 - x + 1)}{(x - 3)(x + 3)} = \dfrac{(-3)(21)}{(-7)(-1)} = \dfrac{-63}{7} = -9 \lt 0 \]
Interval 2: \( x = -2 \) \[ \dfrac{(-1)(7)}{(-5)(1)} = \dfrac{-7}{-5} = \dfrac{7}{5} > 0 \]
Interval 3: \( x = 0 \) \[ \dfrac{(1)(1)}{(-3)(3)} = \dfrac{1}{-9} \lt 0 \]
Interval 4: \( x = 4 \) \[ \dfrac{(5)(13)}{(1)(7)} = \dfrac{65}{7} > 0 \]
Step 5: Include or exclude the critical points
\( x = -3 \) and \( x = 3 \) make the denominator zero and are therefore excluded.
\( x = -1 \) makes the numerator zero which does not satisfy \( \lt 0 \) and is therefore excluded.
Finally, the solution set of the given inequality is: \[ (-\infty, -3) \cup (-1, 3) \]
