Chapter 3 Inequalities in one Variable

Section 3.3 Absolute Value Inequalities and Quadratic Inequalities

3.3.2 Quadratic Absolute Value Inequalities

Info 3.3.4
An inequality is called quadratic in x if it can be transformed into x2 +px+q<0. (Other comparators are allowed as well.)
Hence, quadratic inequalities can be solved in two ways: by investigating the roots and the orientation behaviour of the polynomial (i.e., whether the parabola opens upwards or downwards) and by completing the square. Often completing the square is simpler:
Info 3.3.5
To solve an inequality by completing the square one tries to transform it into the form (x+a )2 <b. Taking the square root then results in the absolute value inequality |x+a|<b with the solution set ]-a-b;-a+b[ if b0. If b<0, the inequality is unsolvable.
The inverted inequality |x+a|>b has the solution set ]-;-a-b[ ]-a+b;[ . For and the corresponding boundary points have to be included.

Always note the calculation rule x2 =|x| described in Module 1.
Example 3.3.6
Find the solution of the inequality 2 x2 4x+2. Collecting the terms on the left-hand side and dividing the inequality by 2 results in x2 -2x-10. Completing the square on the left-hand side to the second binomial formula results in the equivalent inequality x2 -2x+12 or (x-1 )2 2, respectively. Taking the square root results in the absolute value inequality |x-1|2 with the solution set L= ]-;1-2][1+2;[ .

On the other hand, the inequality x2 -2x-10 can be investigated as follows: The left-hand side describes a parabola opened upwards. The roots x1,2 =1±2 can be found using the pq formula:
Since the parabola opens upwards, the inequality x2 -2x-10 is satisfied by the values of x in the parabola branches left and right to the roots, i.e. by the set L= ]-;1-2][1+2;[ .
Info 3.3.7
Depending on the roots of x2 +px+q, the orientation of the parabola and the comparator, the quadratic inequality x2 +px+q<0 (including other comparators) has one of the following solution sets:
  • the set of real numbers ,
  • two branches ]-; x1 [ ] x2 ;[ (including the boundary points for and ),
  • an interval ] x1 ; x2 [ (including the boundary points for and if applicable),
  • a single point x1 ,
  • the pointed set { x1 },
  • the empty set {}.

Fill in the blanks in the following text describing the solution of a quadratic inequality by investigating the behaviour of the parabola:
Exercise 3.3.8
Find the solution set of the inequality x2 +6x<-5. Transformation results in the inequality
<0. Using the pq formula one obtains the set of roots
. The left-hand side describes a parabola opening
. It belongs to an inequality involving the comparator <, hence the solution set is L =