Modulus problems by critical points

When an equa­tion has two mod­uli added together, such as $|x + 3| + |2 - 3x| = 5$, or a mod­u­lus mixed with other terms, such as $|3x - 2| + x = 11$, the quick rules $|x| = \pm k$ no longer work. The reli­able tool is the crit­i­cal point method: cut the num­ber line at the points where each mod­u­lus changes its form, solve a plain equa­tion on each piece, and keep only the answers that really lie on that piece. This les­son teaches the method and applies it to equa­tions and inequal­i­ties.

The crit­i­cal point method#

1. Find the crit­i­cal points. Set each expres­sion inside a mod­u­lus equal to zero. These are the points where the expres­sion changes sign, so the mod­u­lus changes its form.
2. Cut the num­ber line at the crit­i­cal points into inter­vals.
3. On each inter­val, decide the sign of each inside expres­sion, and remove the bars: $|T| = T$ if $T \geq 0$, $|T| = -T$ if $T < 0$.
4. Solve the result­ing equa­tion or inequal­ity, which has no mod­uli left.
5. Keep only the com­mon val­ues: an answer counts only if it lies in the inter­val you assumed.
6. Com­bine the answers from all the inter­vals.

Step 5 is the one stu­dents for­get, and it is the step that throws out false answers.

Exam­ple 1#

Solve $|x + 3| + |2 - 3x| = 5$.

Crit­i­cal points. $x + 3 = 0$ gives $x = -3$; $2 - 3x = 0$ gives $\displaystyle x = \frac{2}{3}$. The crit­i­cal points are $-3$ and $\displaystyle \frac{2}{3}$, which cut the line into three inter­vals.

Case (i): $x < -3$. Here $x + 3 < 0$ and $2 - 3x > 0$. So

$$\displaystyle \begin{aligned} -(x + 3) + (2 - 3x) &= 5 \\ -x - 3 + 2 - 3x &= 5 \\ -4x &= 6 \\ x &= -\frac{3}{2} \end{aligned}$$

But $\displaystyle -\frac{3}{2}$ is greater than $-3$, so it does not lie in this inter­val. No solu­tion here.

Case (ii): $\displaystyle -3 \leq x \leq \frac{2}{3}$. Here $x + 3 \geq 0$ and $2 - 3x \geq 0$. So

$$\begin{aligned} (x + 3) + (2 - 3x) &= 5 \\ -2x + 5 &= 5 \\ x &= 0 \end{aligned}$$

$0$ does lie between $-3$ and $\displaystyle \frac{2}{3}$, so $x = 0$ is a solu­tion.

Case (iii): $\displaystyle x > \frac{2}{3}$. Here $x + 3 > 0$ and $2 - 3x < 0$. So

$$\begin{aligned} (x + 3) - (2 - 3x) &= 5 \\ x + 3 - 2 + 3x &= 5 \\ 4x &= 4 \\ x &= 1 \end{aligned}$$

$\displaystyle 1 > \frac{2}{3}$, so $x = 1$ is a solu­tion.

Answer: $x \in \{0, 1\}$. Check $x = 1$: $|4| + |-1| = 5$.

Exam­ple 2#

Solve $|3x - 2| + x = 11$.

Crit­i­cal point. $3x - 2 = 0$ gives $\displaystyle x = \frac{2}{3}$.

Case (i): $\displaystyle x < \frac{2}{3}$. Then $3x - 2 < 0$, so $|3x - 2| = 2 - 3x$:

$$\displaystyle \begin{aligned} 2 - 3x + x &= 11 \\ -2x &= 9 \\ x &= -\frac{9}{2} \end{aligned}$$

$\displaystyle -\frac{9}{2} < \frac{2}{3}$, so this is a solu­tion.

Case (ii): $\displaystyle x \geq \frac{2}{3}$. Then $|3x - 2| = 3x - 2$:

$$\displaystyle \begin{aligned} 3x - 2 + x &= 11 \\ 4x &= 13 \\ x &= \frac{13}{4} \end{aligned}$$

$\displaystyle \frac{13}{4} > \frac{2}{3}$ (com­pare $\displaystyle \frac{39}{12}$ with $\displaystyle \frac{8}{12}$), so this is also a solu­tion.

Answer: $\displaystyle x \in \left\{-\frac{9}{2}, \frac{13}{4}\right\}$.

Inequal­i­ties with the same method#

The method works just as well for inequal­i­ties. In each inter­val you solve an inequal­ity, and then take the over­lap of its answer with the inter­val.

Exam­ple 3#

Solve $\displaystyle |x + 2| - |x - 1| < x - \frac{3}{2}$.

Crit­i­cal points. $x = -2$ and $x = 1$.

Case (i): $x < -2$. Both $x + 2 < 0$ and $x - 1 < 0$:

$$\displaystyle \begin{aligned} (-x - 2) - (1 - x) &< x - \frac{3}{2} \\ -x - 2 - 1 + x &< x - \frac{3}{2} \\ -3 &< x - \frac{3}{2} \\ -\frac{3}{2} &< x \end{aligned}$$

We need $x < -2$ and $\displaystyle x > -\frac{3}{2}$ together, which is impos­si­ble. No solu­tion.

Case (ii): $-2 \leq x < 1$. Now $x + 2 \geq 0$ and $x - 1 < 0$:

$$\displaystyle \begin{aligned} (x + 2) - (1 - x) &< x - \frac{3}{2} \\ 2x + 1 &< x - \frac{3}{2} \\ x &< -\frac{5}{2} \end{aligned}$$

But $x \geq -2$ in this case, and $\displaystyle -\frac{5}{2} < -2$. No over­lap, no solu­tion.

Case (iii): $x \geq 1$. Both expres­sions are non-neg­a­tive:

$$\displaystyle \begin{aligned} (x + 2) - (x - 1) &< x - \frac{3}{2} \\ 3 &< x - \frac{3}{2} \\ \frac{9}{2} &< x \end{aligned}$$

Every $\displaystyle x > \frac{9}{2}$ also sat­is­fies $x \geq 1$.

Answer: $\displaystyle x \in \left(\frac{9}{2}, \infty\right)$.

Exam­ple 4#

Solve $x^2 - 5|x| + 6 < 0$.

Crit­i­cal point. $x = 0$.

Case (i): $x < 0$. Then $|x| = -x$:

$$x^2 + 5x + 6 < 0 \implies (x + 2)(x + 3) < 0 \implies x \in (-3, -2)$$

All of $(-3, -2)$ is neg­a­tive, so the com­mon val­ues are $(-3, -2)$.

Case (ii): $x \geq 0$. Then $|x| = x$:

$$x^2 - 5x + 6 < 0 \implies (x - 2)(x - 3) < 0 \implies x \in (2, 3)$$

All of these are non-neg­a­tive, so the com­mon val­ues are $(2, 3)$.

Answer: from both cases, $x \in (-3, -2) \cup (2, 3)$.

Exam­ple 5#

Solve $|x^2 - 4x| < 5$.

Here a sin­gle mod­u­lus is less than a num­ber, so use $|T| < a \iff -a < T < a$:

$$-5 < x^2 - 4x < 5$$

This is two inequal­i­ties, both of which must hold.

First: $x^2 - 4x + 5 > 0$. Here $a = 1 > 0$ and $\Delta = 16 - 20 = -4 < 0$, so the parabola lies above the axis: true for all $x \in \mathbb{R}$.

Sec­ond: $x^2 - 4x - 5 < 0$. Here $\Delta = 16 + 20 = 36$, and the roots are $\displaystyle \frac{4 \pm 6}{2}$, that is, $-1$ and $5$. So $x \in (-1, 5)$.

Answer: the com­mon val­ues are $x \in (-1, 5)$.

Prac­tice#

1. Solve $|x - 1| + |x - 3| = 4$.
2. Solve $|2x - 1| + x = 5$.
3. Solve $|x + 1| + |x - 2| = 3$.
4. Solve $x^2 - 4|x| + 3 < 0$.
5. Solve $x^2 - 3|x| - 4 = 0$.
6. Solve $|x^2 - 2x| < 3$.
7. Solve $|x - 2| - |x + 1| = 1$.
8. Solve $|x - 1| + |x + 2| \leq 5$.

Answers#