#### Chapter 2 Equations in one Variable

**Section 2.2 Absolute Value Equations**

# 2.2.3 Mixed Equations

##### **Info 2.2.19 **

If an equation contains both absolute values and other expressions, the case analysis has to be done according to the absolute value terms and applied only to these.

Finally, keep in mind to cross-check the solution sets you found with the case conditions.

##### **Example 2.2.20 **

Solve the equation $|x-1|+{x}^{2}=1$. Here, the case analysis is as follows:

- For $x\ge 1$, the absolute value bars can be replaced by normal brackets which results in the quadratic equation $(x-1)+{x}^{2}=1$ that is transformed into the equation ${x}^{2}+x-2=0$. Using the $pq$ formula we get the two solutions

$\begin{array}{ccc}\multicolumn{1}{c}{{x}_{1}}& =\hfill & -\frac{1}{2}-\sqrt{\frac{9}{4}}\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}-2\hspace{0.5em},\hfill \\ \multicolumn{1}{c}{{x}_{2}}& =\hfill & -\frac{1}{2}+\sqrt{\frac{9}{4}}\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}1\hspace{0.5em}\hfill \end{array}$

of which only ${x}_{2}$ satisfies the case condition.

- For $x<1$, one obtains the quadratic equation $-(x-1)+{x}^{2}=1$ that is transformed into the equation ${x}^{2}-x=0$ or $x\xb7(x-1)=0$, respectively. The product representation indicates the two solutions ${x}_{3}=0$ and ${x}_{4}=1$. Because of the case condition only ${x}_{3}=0$ is a solution of the initial equation.

##### **Exercise 2.2.21 **

Find the solution set of the mixed equation $|x-3|\xb7x=9$.

So, altogether the solution set is $L$$\hspace{0.5em}=\hspace{0.5em}$ .

- If $x$ is in the interval

the absolute value term is non-negative.

One obtains the quadratic equation

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

The solution set is .

Only the solution

satisfies the case condition.

- If $x$ is in the interval

the absolute value term is negative.

One obtains the normalised quadratic equation

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

The solution set is .

`infty`. Do not use the notation $]a;b[$ fir open intervals. Sets can be entered by listing the elements in the form $\{1;2;3\}$. For the set brackets enter AltGr+7 or AltGr+0, respectively.So, altogether the solution set is $L$$\hspace{0.5em}=\hspace{0.5em}$ .

##### **Exercise 2.2.22 **

Find the solutions of the mixed absolute value equation $3|2x+1|=|x-5|$ by visualising the different cases on the number line and finally solving the equation by case analysis. First, visualise the case analysis for each absolute value.

The solution set is .

The solution set is .