Absolute Value Equations

Absolute value equations are equations that contain an absolute value sign.

Example 1. Solve the following equation: ${\left|{x}\right|}=5$.

We know, that absolute value makes any number positive.

What are numbers, that when made positive will give $5$?

They are $5$ and $-5$.

So, $5$ and $-5$ are the roots of the equation.

How to solve absolute value equation ${\left|{x}\right|}={m}$?

• It has two roots: ${m}$ and $-{m}$, if ${m}$ is a positive number.
• It has one root 0, if ${m}={0}$.
• It has no roots, if ${m}$ is a negative number (absolute value can't be negative).

In more complex examples, we need to isolate the absolute value and then just use the above rule.

Example 2. Solve the following equation: ${\left|{x}+{3}\right|}={7}$.

Absolute value is already isolated, so we have 2 cases:

1. ${x}+{3}={7}$, which gives ${x}={4}$.
2. ${x}+{3}=-{7}$, which gives ${x}=-{10}$.

So, ${4}$ and $-{10}$ are roots of the equation.

Let's solve a bit more complex equation.

Example 3. Solve the following equation: ${\left|{2}{x}+{5}\right|}-{8}={14}$.

First, isolate absolute value.

For this add ${8}$ to both sides of the equation: ${\left|{2}{x}+{5}\right|}={22}$.

Absolute value is isolated, so we have 2 cases:

1. ${2}{x}+{5}={22}$, which gives ${x}=\frac{{17}}{{2}}$.
2. ${2}{x}+{5}=-{22}$, which gives ${x}=-\frac{{27}}{{2}}$.

So, $\frac{{17}}{{2}}$ and $-\frac{{27}}{{2}}$ are roots of the equation.

Until now, we considered equations, where variable is only inside absolute value bars. Let's consider equation, where variable is both inside and outside of absolute value bars.

Example 4. Solve the equation ${\left|{2}{x}+{3}\right|}={5}-{x}$.

We can use above rule, but carefully.

Indeed, rule requires the right side to be positive number, but we don't know whether ${5}-{x}$ is positive or negative.

So, after finding potential roots, we need to check them.

Thus, we have two options:

1. ${2}{x}+{3}={5}-{x}$, which gives ${x}=\frac{{2}}{{3}}$.
2. ${2}{x}+{3}=-{\left({5}-{x}\right)}$, which gives ${x}=-{8}$.

We've got two potential solutions, now need to check them.

1. ${\left|{2}\cdot\frac{{2}}{{3}}+{3}\right|}={5}-\frac{{2}}{{3}}$

${\left|\frac{{7}}{{3}}\right|}=\frac{{7}}{{3}}$

$\frac{{7}}{{3}}=\frac{{7}}{{3}}$ (OK)

2. ${\left|{2}\cdot{\left(-{8}\right)}+{3}\right|}={5}-{\left(-{8}\right)}$

${\left|-{13}\right|}={5}+{8}$

${13}={13}$ (OK)

So, $\frac{{2}}{{3}}$ and $-{8}$ are roots of the equation.

Now, let's see an example, when potential solution is not a root.

Example 5. Solve the equation ${\left|{x}+{5}\right|}={2}{x}+{1}$.

As in example 4 we can use above rule, but carefully.

Indeed, rule requires the right side to be positive number, but we don't know whether ${2}{x}+{1}$ is positive or negative.

So, after finding potential roots, we need to check them.

Thus, we have two options:

${x}+{5}={2}{x}+{1}$, which gives ${x}={4}$.

${x}+{5}=-{\left({2}{x}+{1}\right)}$, which gives ${x}=-{2}$.

We've got two potential solutions, now need to check them.

1. ${\left|{4}+{5}\right|}={2}\cdot{4}+{1}$

${\left|{9}\right|}={9}$

${9}={9}$ (OK)

2. ${\left|-{2}+{5}\right|}={2}\cdot{\left(-{2}\right)}+{1}$

${\left|{3}\right|}=-{3}$

${3}\ne-{3}$ (NOT OK)

So, the only root of the equation is ${4}$.

Finally, equation can contain two absolute values.

Example 6. Solve the equation ${\left|{2}{x}+{3}\right|}={\left|{3}{x}+{9}\right|}$.

It seems, that above rule won't work here. But we don't need it here.

What is absolute value? It makes any number positive.

So, absolute value expressions are equal, when quantities inside absolute value bars are equal or equal with different signs.

We again have two cases here:

1. ${2}{x}+{3}={3}{x}+{9}$, which gives ${x}=-{6}$
2. ${2}{x}+{3}=-{\left({3}{x}+{9}\right)}$, which gives ${x}=-\frac{{12}}{{5}}$

Thus, equation has two roots: $-{6}$ and $-\frac{{12}}{{5}}$.

Now it is time to exercise.

Exercise 1. Find roots of the equation ${\left|{x}+{2}\right|}={5}$.

Answer: ${3}$ and $-{7}$.

Exercise 2. Find roots of the equation ${\left|{2}{x}\right|}={7}$.

Answer: $\frac{{7}}{{2}}$ and $-\frac{{7}}{{2}}$.

Exercise 3. Find roots of the equation ${\left|{2}{x}+{1}\right|}={0}$.

Answer: $-\frac{{1}}{{2}}$.

Exercise 4. Solve the following equation: ${\left|{3}{x}+{5}\right|}+{8}={7}$.

Exercise 5. Solve the following equation: ${\left|{3}{x}+{7}\right|}={12}$.

Answer: $\frac{{5}}{{3}}$ and $-\frac{{19}}{{3}}$.

Exercise 6. Solve the following equation: ${\left|{4}{x}+{8}-{2}{\left({x}+{5}\right)}\right|}-{1}={15}$.

Answer: ${9}$ and $-{7}$.

Exercise 7. Solve the following equation: ${\left|\frac{{{2}{\left({x}+{2}\right)}-{5}}}{{3}}\right|}+{3}={22}$.

Answer: ${29}$ and $-{28}$.

Exercise 8. Solve the following equation: ${\left|{2}{x}+{8}\right|}={1}-{2}{x}$.

Answer: $-\frac{{7}}{{4}}$.

Exercise 9. Solve the following equation: ${\left|{4}{x}+{5}\right|}={3}{x}+{1}$.

Exercise 10. Solve the following equation: ${\left|{5}{x}+{10}\right|}={1}-{x}$.
Answer: $-\frac{{11}}{{4}}$ and $-\frac{{3}}{{2}}$.
Exercise 11. Solve the following equation: ${\left|{3}{x}+{7}\right|}={\left|{4}{x}-{5}\right|}$.
Answer: ${12}$ and $-\frac{{2}}{{7}}$.