Absolute Value Equations

Absolute value equations are equations involving expressions with the absolute value functions. This wiki intends to demonstrate and discuss problem solving techniques that let us solve such equations.

A very basic example would be as follows:

Find all values of $x$ satisfying $|x-2| + |x-4| = 4.$

Usually, the basic approach is to analyze the behavior of the function before and after the point where they reach 0. For example, for $|x-a|$ one could analyze the cases where $x > a$ or $x < a$, or even $x = a$ if required. However, these problems are often simplified with a more sophisticated approach like being able to eliminate some of the cases, or graphing the functions. In this wiki, we intend to discuss this techniques along with strategies on when to use which.

• Introduction to absolute value equations

• Methodology to solve absolute value equations: the techniques used to solve absolute value equations and when to use which one

Take an example to describe the following methodology:

1) Understanding absolute value--positive, negative case (or graph approach)
2) Determining possible solutions
3) Verifying solutions

Explain - How do we use this technique to solve absolute value equations?

• Remember to verify the possible solutions - why and how?

• 2-3 examples in increasing order of difficulty - explaining how we squared both sides to solve more difficult problems

• Followed by 1-2 TIY problems - relevant to be solved by case work technique

Suppose we have an equation of the form $\lvert a \rvert = \lvert b \rvert$. Since both sides are positive, we can square them without adding extraneous solutions: $$a^2=b^2.$$ Then solve it as an ordinary equation: $$\begin{aligned} a^2-b^2&=0 \\ (a+b)(a-b)&=0. \end{aligned}$$ So we see that $a=-b$ or $a=b$.

Solve the equation $\lvert 3x+4 \rvert = \lvert 2x-7 \rvert$ for real $x$.

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We square both sides to obtain $$(3x+4)^2=(2x-7)^2.$$ Here we don't need to expand both sides; just apply the difference of two squares to find the factors: $$\begin{aligned} (3x+4+2x-7)(3x+4-2x+7)&=0 \\ (5x-3)(x+11)&=0. \end{aligned}$$ The solutions are $\left\{\frac{3}{5}, -11\right\}. \ _ \square$

Because absolute value can be defined as piecewise functions, depending on where the value of $x$ is with respect to the number line, you have to work with a different "piece" of the piecewise function.

General steps:

1. Using the definition of absolute value as a piecewise function, "undo" the absolute value sign(s) and write cases. For example, we know that the expression in the absolute value sign can either be positive or negative.

2. Solve each case for $x$.

3. Verify the solutions.

Find all real values of $x$ such that $| 3x - 4 | - 2 = 3.$

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We first isolate the absolute value onto one side:

$$\begin{aligned} | 3x - 4 | - 2 & = 3\\ | 3x - 4 | &= 5. \end{aligned}$$

Now, we "undo" the absolute value signs and split the equation into its two cases, the positive case and the negative case:

$$\begin{array}{rlcccrl} (3x - 4) &= 5 &&\text{ or } && -(3x - 4) &= 5\\ 3x - 4 &= 5 &&\text{ or } && -3x + 4 &= 5\\ 3x &= 9 &&\text{ or } && -3x &= 1\\ x &= 3 &&\text{ or } && x &= -\frac{1}{3}. \ _\square \end{array}$$

Find all real values of $x$ such that $|x+1| + |2x+3| = 5$.

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There are four possible cases, but one will be eliminated due to impossibility:

Case 1. If $x+1$ and $2x+3$ are both positive, then
$$\begin{aligned} x+1 + 2x+3 &= 5 \\ 3x + 4 &= 5 \\ 3x &= 1 \\ x &= \dfrac{1}{3}. \end{aligned}$$

Case 2. If $x+1$ is negative and $2x + 3$ is positive, then $$\begin{aligned} - x - 1 + 2x + 3 &= 5 \\ x + 2 &= 5 \\ x &= 3. \end{aligned}$$ However, when $x = 3$, $x+1$ and $2x+3$ are both positive, so this is not a valid solution to the equation.

Case 3. If $x+1$ and $2x+3$ are both negative, then $$\begin{aligned} -x - 1 - 2x - 3 &= 5 \\ -3x - 4 &= 5 \\ -3x &= 9 \\ x &= -3. \end{aligned}$$

Case 4. If $x +1$ is positive and $2x + 3$ is negative, it is an impossible case. Graph the two lines if you are not convinced.

Therefore, the solution set is $\left \{ -3, \frac{1}{3} \right \}.\ _\square$

Find all real values of $x$ such that

$$|x+2|+|2x+6|+|3x-3|=12.$$

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In this problem we are dealing with 3 terms of absolute values. Their turning points (the values of $x$ such that they change sign) of the three terms are $x=-2, x=-3, x=1,$ respectively. Hence, we need to check the cases $-\infty < x \leq -3$, $-3<x\leq -2$, $-2 < x \leq 1$, $1<x<\infty$.

Case 1. $\, -\infty < x \leq -3$
In this case, the three terms will always be negative. Hence, $$\begin{aligned} -(x+2)-(2x+6)-(3x-3)&=12 \\ x &= -\frac{17}{6}. \end{aligned}$$ However, $x=-\frac{17}{6} >-3$ is not within the domain $-\infty < x \leq -3$. Thus this solution is not valid.

Case 2. $\, -3<x\leq -2$
In this case, the three terms will be negative, positive, and negative, respectively. Hence, $$\begin{aligned} -(x+2)+(2x+6)-(3x-3)&=12 \\ x &= -\frac{5}{2}. \end{aligned}$$ $x=-\frac{5}{2}$ lies between $-3$ and $-2$. Thus $\boxed{x=-\frac{5}{2}}$ is one of the solutions.

Case 3. $\, -2 < x \leq 1$
In this case, the three terms will be positive, positive, and negative, respectively. However, $$\begin{aligned} (x+2)+(2x+6)-(3x-3)=11 \neq 12. \end{aligned}$$ Thus there is no solution within this domain.

Case 4. $\, 1<x<\infty$
In this case, the three terms are always positive. Hence, $$\begin{aligned} (x+2)+(2x+6)+(3x-3)&=12 \\ x &= \frac{7}{6}, \end{aligned}$$ which lies between $1$ and $\infty$. Thus $\boxed{x=\frac{7}{6}}$ is another solution.

In conclusion, $x=-\frac{5}{2}$ and $x=\frac{7}{6}$ are the solutions for the given equation. $_\square$

Find all real values of $x$ such that $|x||x+1| = 2$.

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Case 1. $\, x, x+1$ both positive
$$\begin{aligned} x(x+1)-2 &= 0 \\ x^2 +x - 2 &= 0 \\ x &= 1, x = -2. \end{aligned}$$ Reject $x = -2$ because it does not make both $x$ and $x +1$ positive.

Case 2. $\, x$ negative, $x + 1$ positive
$$\begin{aligned} -x(x+1)-2 &= 0 \\ -x^2 - x - 2 &= 0 \\ x^2 + x + 2 &= 0 \\ x &= \dfrac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 2 }}{2}. \end{aligned}$$ We only asked for real solutions, so at this point we ignore this case because we're going to get imaginary results.

Case 3. $\, x$ positive, $x +1$ negative
This is an impossible case (graph the lines and you'll see why), so we can ignore it.

Case 4. $\, x, x+1$ both negative
Because they're both negative, the negatives end up "canceling" and become positive, which was what we had in Case 1. However, the restriction is different from Case 1 (here, both $x$ and $x +1$ have to be negative, not positive ), so instead of rejecting $x = -2$, we reject $x = 1$ from this case. Basically, in this specific case 4, $x = 1$ is not a possible solution, but it does not mean it's not a possible solution for Case 1 because we're simply going piece by piece in this piecewise function--in the end we will take the union of all possible solutions.

Thus, the solutions are $\left \{ -2, 1 \right \}$. $_\square$

Sometimes absolute value equations have a ridiculous number of cases and it would take too long to go through every single case. Therefore, we can instead graph the absolute value equations using the definition of absolute value as a piecewise function. To get each piece, you must figure out the domain of each piece. This method is highly beneficial when the question writer asks for the number of solutions instead of the actual solutions. Let's work through some examples to see how this is done.

Find all real solutions to $|3x-4| = 5$.

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To graph this, there are two possible cases: when $3x - 4$ is positive, and when $3x-4$ is negative.

When is $3x-4$ positive? $$\begin{aligned} 3x - 4 &> 0 \\ 3x &> 4 \\ x &> \dfrac{4}{3}. \end{aligned}$$ (Also, when $x < \frac{4}{3}$, $3x- 4$ will be negative.)

We know that there will be a "turning point" at $x = \frac{4}{3}$ for the graph of $y = |3x-4|$.

Finally, using the definition of absolute value, we know that when $x > \frac{4}{3}$, $y = 3x - 4$, and when $x \leqslant \frac{4}{3}$, $y = -3x + 4$. We now just need to graph $y = 5$ and look for the intersections.

You can see that the solutions are $\left \{ -\frac{1}{3} , 3 \right \}.\ _\square$.

Another benefit of this graphing technique is that you do not need to verify any of the solutions--since we are only graphing the pieces that are actually mathematically possible, we get all the solutions we are looking for, no less and no more. If you could not discern the solutions from the picture, you can simply solve the equation for each case.

Find all real solutions to $|x+1| + |2x+3| = 5$.

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The possible cases are that
$\hspace{0.5cm}$1. $\, x+1, 2x+3$ are both positive;
$\hspace{0.5cm}$2. $\, x+1$ is negative and $2x+3$ is positive;
$\hspace{0.5cm}$3. $\, x+1 , 2x + 3$ are both negative.
We need to figure out the domains for which each of these holds.

Case 1 holds when $x > -1$.
Case 2 holds when $-\frac{3}{2} < x< -1$.
Case 3 holds when $x < -\frac{3}{2}$.

Now, let's write our piecewise function.

When $x > -1$, we have $y = x+1 + 2x + 3 = 3x + 4$.
When $-\dfrac{3}{2} < x< -1$, we have$y = -x -1 + 2x + 3 = x + 2$.
When $x < - \dfrac{3}{2}$, we have $y = -x - 1 -2x - 3 = -3x -4$.

As you can see in the graph, the solutions for the given equation are $\left \{ -3, \frac{1}{3} \right \}.\ _\square$ .

Find all real solutions to $|x||x+1| = 2$.

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To graph this, we again only look at the possible cases and when they would occur:
$\hspace{0.5cm}$1. $\, x, x+1$ both positive
$\hspace{0.5cm}$2. $\, x$ negative, $x+1$ positive
$\hspace{0.5cm}$3. $\, x, x+1$ both negative.

Case 1 is true when $x>0$.
Case 2 is true when $-1 < x < 0$.
Case 3 is true when $x<-1$.

When $x>0$ and when $x< -1$, we have $y = x(x+1) = x^2 + x$.
When $-1 < x < 0$, we have $y = -x(x+1) = -x^2 - x$.

It is evident that the solutions are $\{-2, 1\}.\ _\square$

• Any other technique (fact, definition) you can use to solve the problems? Otherwise move on to the followings.

• 3-4 examples solved by using a mix of more than one of above techniques

• Add guiding text in between. Guiding text means phrasing the section in a way that it keeps on telling the reader what's going on in this section.

• 3-4 TIY problems - using multiple techniques to solve

What is the sum of all real numbers $x$ satisfying $$x^2-\sqrt{x^2} = \lvert x-1 \rvert +5?$$

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Observe that $\sqrt{x^2}=\lvert x \rvert.$ Then the given equation becomes $$x^2-\lvert x \rvert= \lvert x-1 \rvert +5.$$

• If $x<0,$ then we rewrite the equation to obtain $$\begin{aligned} x^2-(-x)&=-(x-1)+5\\ x^2+2x-6&=0\\ x&=-1\pm \sqrt{7}\\ x&=-1-\sqrt{7}. \qquad (\text{since } x<0) \end{aligned}$$

• If $0\le x<1,$ then we rewrite the equation to obtain $$\begin{aligned} x^2-x&=-(x-1)+5\\ x^2&=6\\ x&=\pm \sqrt{6}, \end{aligned}$$ which do not satisfy the assumption $0\leq x<1.$ Thus there are no solutions in this interval.

• If $x\ge 1,$ then we rewrite the equation to obtain $$\begin{aligned} x^2-x&=x-1+5\\ x^2-2x-4&=0\\ x&=1\pm \sqrt{5}\\ x&=1+\sqrt{5}. \qquad (\text{since } x\ge 1) \end{aligned}$$

Therefore, the above three cases give two solutions $x=-1-\sqrt{7}$ and $x=1+\sqrt{5},$ the sum of which is $\sqrt{5}-\sqrt{7}.$ $_\square$

[IMO 1959/2] Solve the equation $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A$ for real $x$ (where the square roots are only defined for non-negative values), when

• $A=\sqrt{2}$;
• $A=1$;
• $A=2$.

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Here we don't see any absolute value involved with the equation. Before doing anything, note that our first restriction for $x$ is $x \geq \frac{1}{2}$ and for $A$ is $A>0$. Intuitively, we could square both sides to get rid of some square roots: $$\begin{aligned} x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{x^2-(2x-1)}&=A^2 \\ 2x+2\sqrt{(x-1)^2}&=A^2. \end{aligned}$$ Great! We find a perfect square inside the square root, so an absolute value will appear: $$2x+2|x-1|=A^2.$$ Now we are going to find the possible cases for $A$:

• When $x-1 > 0$, we have $$\begin{aligned} 2x+2(x-1)&=A^2 \\ x&=\dfrac{A^2+2}{4}. \end{aligned}$$ Then, by our assumption of $x-1 > 0$, we get that this solution only works when $A^2 > 2$.

• When $x-1 \leq 0$, something interesting happens: $$\begin{aligned} 2x-2(x-1)&=A^2 \\ 2&=A^2. \end{aligned}$$

So, when $A^2=2 (\text{or }A=\sqrt{2})$, the equation becomes independent of $x$, implying that any value of the interval $x \in \left[\frac{1}{2},1\right]$ will be a solution for the first point.

When $A=1$, there are no solutions by our restriction of $A^2 \geq 2$.

Finally, when $A=2$ we have $$x=\dfrac{2^2+2}{4}=\frac{3}{2}.\ _\square$$

What happens if we allow the square roots to admit negative values?

Sometimes, in minimization problems, it often helps us to see that the value of an expression inside the absolute value is at least 0.