#### Chapter 3 Inequalities in one Variable

**Section 3.3 Absolute Value Inequalities and Quadratic Inequalities**

# 3.3.2 Quadratic Absolute Value Inequalities

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 $|x+a|<\sqrt{b}$ with the solution set $\text{}]-a-\sqrt{b};-a+\sqrt{b}[\text{}$ if $b\ge 0$. If $b<0$, the inequality is unsolvable.

The inverted inequality $|x+a|>\sqrt{b}$ has the solution set $\text{}]-\infty ;-a-\sqrt{b}[\text{}\cup \text{}]-a+\sqrt{b};\infty [\text{}$. 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 $(x-1{)}^{2}\ge 2$, respectively. Taking the square root results in the absolute value inequality $|x-1|\ge \sqrt{2}$ with the solution set $L=\text{}]-\infty ;1-\sqrt{2}]\cup [1+\sqrt{2};\infty [\text{}$.

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 ${x}_{\mathrm{1,2}}=1\pm \sqrt{2}$ 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 $L=\text{}]-\infty ;1-\sqrt{2}]\cup [1+\sqrt{2};\infty [\text{}$.

##### **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 $\mathbb{R}$,

- two branches $\text{}]-\infty ;{x}_{1}[\text{}\cup \text{}]{x}_{2};\infty [\text{}$ (including the boundary points for $\le $ and $\ge $),

- an interval $\text{}]{x}_{1};{x}_{2}[\text{}$ (including the boundary points for $\le $ and $\ge $ if applicable),

- a single point ${x}_{1}$,

- the pointed set $\mathbb{R}\setminus \{{x}_{1}\}$,

- 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 ${x}^{2}+6x<-5$. 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$$\hspace{0.5em}=\hspace{0.5em}$

.

$<0$. 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$$\hspace{0.5em}=\hspace{0.5em}$

.