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 b≥0. 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-1≥0. Completing the square on the left-hand side to the second binomial formula results in the equivalent inequality x2 -2x+1≥2 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-1≥0 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-1≥0 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 ℝ,
  • 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 ℝ∖{ 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  = 
.  


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