#### 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 ${x}^{2}+px+q<0$. (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 $\left(x+a{\right)}^{2}. Taking the square root then results in the absolute value inequality $|x+a|<\sqrt{b}$ with the solution set $\text{}\right]-a-\sqrt{b};-a+\sqrt{b}\left[\text{}$ if $b\ge 0$. If $b<0$, the inequality is unsolvable.
The inverted inequality $|x+a|>\sqrt{b}$ has the solution set $\text{}\right]-\infty ;-a-\sqrt{b}\left[\text{}\cup \text{}\right]-a+\sqrt{b};\infty \left[\text{}$. For $\le$ and $\ge$ the corresponding boundary points have to be included.

Always note the calculation rule $\sqrt{{x}^{2}}=|x|$ described in Module 1.
##### Example 3.3.6
Find the solution of the inequality $2{x}^{2}\ge 4x+2$. Collecting the terms on the left-hand side and dividing the inequality by $2$ results in ${x}^{2}-2x-1\ge 0$. Completing the square on the left-hand side to the second binomial formula results in the equivalent inequality ${x}^{2}-2x+1\ge 2$ or $\left(x-1{\right)}^{2}\ge 2$, respectively. Taking the square root results in the absolute value inequality $|x-1|\ge \sqrt{2}$ with the solution set $L=\text{}\right]-\infty ;1-\sqrt{2}\right]\cup \left[1+\sqrt{2};\infty \left[\text{}$.

On the other hand, the inequality ${x}^{2}-2x-1\ge 0$ can be investigated as follows: The left-hand side describes a parabola opened upwards. The roots ${x}_{1,2}=1±\sqrt{2}$ can be found using the $\mathrm{pq}$ formula: Since the parabola opens upwards, the inequality ${x}^{2}-2x-1\ge 0$ is satisfied by the values of $x$ in the parabola branches left and right to the roots, i.e. by the set $L=\text{}\right]-\infty ;1-\sqrt{2}\right]\cup \left[1+\sqrt{2};\infty \left[\text{}$.
##### Info 3.3.7

Depending on the roots of ${x}^{2}+px+q$, the orientation of the parabola and the comparator, the quadratic inequality ${x}^{2}+px+q<0$ (including other comparators) has one of the following solution sets:
• the set of real numbers $ℝ$,
• two branches $\text{}\right]-\infty ;{x}_{1}\left[\text{}\cup \text{}\right]{x}_{2};\infty \left[\text{}$ (including the boundary points for $\le$ and $\ge$),
• an interval $\text{}\right]{x}_{1};{x}_{2}\left[\text{}$ (including the boundary points for $\le$ and $\ge$ if applicable),
• a single point ${x}_{1}$,
• the pointed set $ℝ\setminus \left\{{x}_{1}\right\}$,
• the empty set $\left\{\right\}$.

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 ${x}^{2}+6x<-5$. 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$$=$
.