Step-by-Step Guide to Solving Polynomial and Rational
Inequalities with Examples

Learn how to solve polynomial and rational inequalities with this step-by-step tutorial. This guide includes clear explanations and detailed example problems to help students master the techniques for analyzing inequality expressions.

Example 1

Solve the inequality:

\[ 2 - \dfrac{7 x}{5} \gt -x + 3 \]
View Solution and Explanation
  • Multiply all terms by 5 and simplify: \[ 10 - 7x > -5x + 15 \]
  • Move all \(x\) terms to one side: \[ 10 > 2x + 15 \]
  • Subtract 15 from both sides: \[ -5 > 2x \]
  • Divide by 2: \[ -\frac{5}{2} > x \quad \Rightarrow \quad x < -\frac{5}{2} \]
  • Final Answer (Interval Form): \[ \left(-\infty,\ -\frac{5}{2}\right) \]

Example 2

Solve the polynomial inequality:

\[ -4 x^2 \gt 4 x - 8 \]
View Solution and Explanation
  • Move all terms to one side: \[ -4x^2 - 4x + 8 > 0 \]
  • Multiply both sides by \(-1\) (reversing the inequality sign): \[ 4x^2 + 4x - 8 < 0 \]
  • Divide by 4 and solve the associated equation: \[ x^2 + x - 2 = 0 \quad \Rightarrow \quad (x + 2)(x - 1) = 0 \]
  • Find where the expression is zero (critical points):
    \(x + 2 = 0 \implies x = -2\)
    \(x - 1 = 0 \implies x = 1\)
  • Analyze intervals determined by critical points: \((-\infty, -2)\), \((-2, 1)\), and \((1, \infty)\).
  • Test sign in each interval:
    • For \(x = -3\) in \((-\infty, -2)\): \((x + 2)(x - 1)\) is positive.
    • For \(x = 0\) in \((-2, 1)\): \((x + 2)(x - 1)\) is negative.
    • For \(x = 2\) in \((1, \infty)\): \((x + 2)(x - 1)\) is positive.
  • Since we want the expression to be less than 0, we choose the negative interval:
    Final Answer: \((-2,\ 1)\)

Example 3

Solve the rational inequality:

\[ \dfrac{1}{x+4} - \dfrac{2}{x-3} \ge 0 \]
View Solution and Explanation
  • Get a common denominator: \[ \frac{(x - 3) - 2(x + 4)}{(x + 4)(x - 3)} \ge 0 \]
  • Simplify the numerator: \[ \frac{-x - 11}{(x + 4)(x - 3)} \ge 0 \]
  • Multiply numerator and denominator by \(-1\) (reversing the inequality sign): \[ \frac{x + 11}{(x + 4)(x - 3)} \le 0 \]
  • Identify critical points:
    Numerator zero: \(x = -11\)
    Denominator zeros (undefined points): \(x = -4\), \(x = 3\)
  • Test intervals using critical points \(-\infty, -11, -4, 3, \infty\):
    Interval Test Point Sign
    \((-\infty, -11)\) \(x = -12\) \(\mathbf{-}\)
    \((-11, -4)\) \(x = -5\) \(\mathbf{+}\)
    \((-4, 3)\) \(x = 0\) \(\mathbf{-}\)
    \((3, \infty)\) \(x = 4\) \(\mathbf{+}\)
  • Choose intervals where the expression is less than or equal to 0: \[ (-\infty, -11] \cup (-4, 3) \]

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