Absolute Value Equations

Absolute value as distance

The absolute value of a number is its distance from $0$ on the number line. Distance is never negative, so $|x| \ge 0$ always. Some problems call the absolute value the modulus. It is the same thing.

So $|x-40|=15$ says: "the distance from $x$ to $40$ is $15$." Drag the points and see where they land.

Back to the start: the diner is at $|x-40|=15$, so $x-40=15$ or $x-40=-15$. It is at mile marker $55$ or $25$.

Solving $|expression| = k$

Only two numbers are exactly $k$ units from $0$: $k$ and $-k$. So the inside of the bars can only equal one of those two values.

Slide $k$ and watch the horizontal line cross the graph of $y=|2x-1|$. Two crossings for $k>0$, one at $k=0$, none for $k<0$.

Fractions cause no trouble. For $\left|\dfrac{x}{2}+1\right|=3$, solve $\dfrac{x}{2}+1=3$ and $\dfrac{x}{2}+1=-3$ the usual way. You get $x=4$ and $x=-8$.

No solution, zero, and two absolute values

When the right side is $0$ there is only one case. $|A|=0$ means $A=0$, because $-0=0$ gives nothing new. For example, $|x^2-16|=0$ means $x^2=16$, so $x=4$ or $x=-4$. The two answers come from the quadratic, not from splitting.

Now what if both sides have absolute values? Two numbers have the same absolute value when they are the same distance from $0$. They are either equal or opposites.

Quadratics and absolute value

If the inside is quadratic, each case is a quadratic equation, so you can get up to four real solutions. Use factoring or the quadratic formula on each one.

A second pattern: the equation is quadratic in $|x|$. Since $x^2=|x|^2$, substitute $u=|x|$ and remember $u\ge0$.

How many solutions does $|f(x)|=k$ have? Think of the graph of $|f(x)|$: it's the graph of $f$ with the part below the x-axis flipped up. Here $f(x)=x^2-2x-3=(x-3)(x+1)$ has its vertex at $(1,-4)$, and after flipping it becomes a peak at $(1,4)$. Slide $k$ and count the crossings.

The count depends on $k$ and the peak height $4$. For $k>4$ there are $2$ solutions, for $k=4$ there are $3$, for $0

Variable on both sides: check for extraneous solutions

In $|A(x)|=B(x)$ the right side can change sign, and an absolute value can never equal a negative number. So after solving the two cases, you must make sure $B(x)\ge0$.

Rational equations with absolute value

With a fraction inside the bars, you split into two cases as usual, but you must also exclude values that make the denominator $0$. Any candidate that does is not a solution.

A shortcut: when the numerator and denominator share a factor, you can simplify, but still write down the excluded value. For $\left|\dfrac{x^2-x-2}{x^2+x}\right|=1$, we have $\dfrac{(x-2)(x+1)}{x(x+1)}$, so the excluded values are $x=0$ and $x=-1$ even after canceling.

Two absolute values: solve by intervals

If two absolute values are added or subtracted (not set equal), you can't split into just two cases. Each absolute value changes behavior at a different point. The idea is to cut the number line at those points and remove the bars in each piece.

There is a nice way to see this: $|x-1|+|x-4|$ is the sum of the distances from $x$ to $1$ and to $4$. Between $1$ and $4$ that sum is always $3$, and outside it grows. That's why the middle interval gave "$3=5$".

Nested and combined absolute values

Nested. Peel the absolute values from the outside in, one layer at a time.

Absolute value mixed with other terms. Split by the sign of the expression inside the bars. In each case solve, then keep only solutions that fit that case.

Quadratic equals linear, both in absolute value. The rule $|A|=|B|\iff A=\pm B$ still works, and since you never squared or divided, there is nothing to reject. For example, $|x^2-1|=|x+1|$ gives $x^2-1=x+1$ (so $x=2$ or $x=-1$) and $x^2-1=-x-1$ (so $x=0$ or $x=-1$). The solutions are $x=-1$, $0$, and $2$. A repeated solution counts once.

Applications: tolerance and deviation

Absolute value equations describe "how far from a target." In manufacturing, a part is acceptable if its length is within a tolerance of the target. The two solutions are the upper and lower limits.

Summary