In this post, we’ll look at solving inequalities where one (or more) of the factors are always positive or negative. We’ve talked about how to determine whether a quadratic function is always positive or negative (or neither) in this post here. Look through it, as we’ll need to use this concept for this section.
To look at how to solve such inequalities, we’ll make one side of the inequality 0, and then factorize the other side completely. Then, derive the answers by comparing positive and negative numbers.
Example 1: Solve the inequality (x² +x +4)(x-3) < 0
For x² +x +4 , completing the square gives (x+0.5)² + 3.75. Since x² +x +4 = (x+0.5)² + 3.75≥ 0 for all values of x, x² +x +4 is always positive.
For (x² +x +4)(x-3) < 0, given that x² +x +4 is always positive, x-3< 0 (since positive number multiply by negative number gives a negative number, which is lesser than 0).
x-3 < 0
x< 3
The solution for (x² +x +4)(x-3) < 0 is x <3.
Example 2: Solve the inequality (x-3)/(x² +x +4) ≥ 0
For x² +x +4 , completing the square gives (x+0.5)² + 3.75. Since x² +x +4 = (x+0.5)² + 3.75≥ 0 for all values of x, x² +x +4 is always positive.
For (x-3)/ (x² +x +4) ≥ 0, given that x² +x +4 is always positive, x-3≥ 0 (since positive number multiply by positive number gives a positive number or 0, which is ≥ than 0 ).
Hence x-3 ≥ 0,
x ≥ 3
The solution for (x-3)/ (x² +x +4) ≥ 0 is x ≥ 3.
Learn H2 A Level Math Inequalities
Here are the complete notes for solving inequalities for H2 Math:
- Solving inequalities using the graphic calculator
- Solving inequalities involving polynomials
- Solving inequalities involving polynomial fractions
- Quadratic functions that are always positive or negative
- Solving inequalities with functions that are always positive or negative
- Modulus Functions
- Solving inequalities involving modulus functions 1
- Solving inequalities involving modulus functions 2
- Solving inequalities involving modulus functions 3
All the notes for H2 A Level Math
Go here to find all the notes and resources for H2 A level Math.