Inequalities Involving Quadratic or Rational Functions - SS2 Mathematics Lesson Note

Here we discuss the solution to inequalities of the form:

\(p(x)q(x) \geq 0\) \(\frac{p(x)}{q(x)} \geq 0\) \(p(x)q(x) \leq 0\) \(\frac{p(x)}{q(x)} \leq 0\)

Where \(p(x)\) and \(q(x)\) are linear functions of \(x\).

1. Now if \(p(x)q(x) > 0\) or \(p(x)q(x) = 0\) provided \(q(x) \neq 0\), then both \(p(x)\) and \(q(x)\) must be positive (i.e. \(> 0\)) or both must be negative (i.e. \(< 0\)) for their product/quotient to be positive.

Also, if \(p(x)q(x) < 0\) or \(p(x)q(x) = 0\) provided \(q(x) \neq 0\), then \(p(x)\ > \ 0\) and \(q(x)\ < \ 0\) or \(p(x)\ < \ 0\) and \(q(x)\ > 0\) for their product/quotient to be negative.

Example 3 Solve the inequality \((x - 1)(x + 4) \leq 0\)

Solution

Given \((x - 1)(x + 4) \leq 0\)

Either \(x - 1 \leq 0\) or \(x + 4 \geq 0\)

Then, \(x \leq 1\) or \(x \geq - 4\), that is, \(- 4 \leq x \leq 1\)

OR

Either \(x - 1 \geq 0\) or \(x + 4 \leq 0\)

Then, \(x \geq 1\) or \(x \leq - 4\), and this makes no sense.

Therefore, the solution is \(- 4 \leq x \leq 1\)