Absolute-value equations can be fairly simple when they contain only one absolute-value expression, and if that expression is linear. But these equations can have many more than just one expression in absolute-value bars, and the bars may contain expressions more complicated than mere straight lines. We'll start with a quadratic expression as the argument (that is, as the expression inside the absolute-value bars).
The quadratic inside the absolute-value bars might be positive and might be negative, depending on where I look. This is a positive quadratic, so it's an upward-opening parabola. The quadratic will be negative between the two zeroes (because that's where it will cross below the x-axis). Are there any zeroes?
Content Continues Below
The absolute value is isolated on the left-hand side, so I can drop the bars and split the equation into its two cases:
Case 1 ("plus"):
x^{2} – 4x – 5 = +7
x^{2} – 4x – 12 = 0
(x – 6)(x + 2) = 0
x = 6, x = –2
So if the quadratic is positive, there are two solution values to the equation. What if the quadratic is negative?
Case 2 ("minus"):
x^{2} – 4x – 5 = ▬7
x^{2} – 4x + 2 = 0
x^{2} – 4x + 2 = 0
There are no factors of +2 that sum to –4, so this quadratic does not factor. To solve, I'll have to apply the Quadratic Formula:
So the "minus" case has two solutions, too. Then my complete answer is:
Content Continues Below
When one absolute-value expression is inside another one, this is called a "nested" expression. The solution process works the same as what we've seen before; the difference will be the number of times we have to split things into cases.
This equation has one absolute-value expression nested inside another one. I'll work from the outside in, splitting things into their cases. First, suppose the argument of the outer bars is negative (that is, do the "minus" case):
| 2x + 5 | – 3 = –8
| 2x + 5 | = –5
I have to stop at this point, because that last line above says that an absolute value is equal to a negative number. But absolute values are never negative! So this case doesn't work; it will provide no valid solutions to the original equation.
Now I'll consider the case where the argument of the outer bars is positive (that is, I'll do the "plus" case):
| 2x + 5 | – 3 = 8
| 2x + 5 | = 11
Advertisement
This equation is valid, and I've got the remaining absolute value isolated, so now I'll consider the two cases again. First, I'll look at the "minus" case:
2x + 5 = –11
2x = –16
x = –8
(It's okay to get a "minus" value for the variable; the only restriction is that I can't get a "minus" value for an absolute-value expression.) Now for the "plus" case:
2x + 5 = +11
2x = 6
x = 3
Then my complete answer is:
x = –8, 3
This looks awful, but nested absolute values aren't actually that bad. I just have to work in steps. I'll start by taking the outermost absolute-value bars off. Since the original equation is "equals zero", I don't even have to bother with considering the "plus" and "minus" cases:
||| x – 1 | – 1 | – 1 | – 1 = 0
||| x – 1 | – 1 | – 1 | = 1
Now I need to start considering cases. I'll start with the "minus" for the equation above.
|| x – 1 | – 1 | – 1 = –1
|| x – 1 | – 1 | = 0
This is another "equals zero", so I can go straight to the next set of bars:
| x – 1 | – 1 = 0
| x – 1 | = 1
Now I'll split into cases again; I'll start with the "minus" case:
x – 1 = –1
x = 0
Okay; that's one solution. Now I'll consider the "plus" case:
x – 1 = +1
x = 2
That's another solution.
I'd found these two solutions by taking the first "minus" case, and tracing it down to its end. That consideration started back at the purple "Now", above. So I need to start there, and consider the "plus" case.
|| x – 1 | – 1 | – 1 = +1
|| x – 1 | – 1 | = 2
I'll split this into two cases; I'll consider the "minus" case first:
| x – 1 | – 1 = –2
| x – 1 | = –1
This case doesn't work, because it's telling me that an absolute value is equal to a negative number. That can't be true. So I discard this branch and try the "plus" case:
| x – 1 | – 1 = +2
| x – 1 | = 3
This works, so I can proceed to the two branches for this equation. I'll do the "minus" case first:
x – 1 = –3
x = –2
Now I'll do the "plus" case:
x – 1 = +3
x = 4
I've considered every case, and have arrived at four solutions. My complete answer is:
x = –3, 0, 2, 4
Affiliate
By the way, you may find it helpful, when you split an absolute value into its two cases, to write them side-by-side, and work down the page in columns. That's what I did in my scratch-work, when computing the answers above. (The columns don't work well on web pages, is the reason for how I've formatted things above.) But if you use multiple columns, be sure to label your hand-in work clearly, so the grader can follow what you're doing.
You can use the Mathway widget below to practice solving an absolute-value equation. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's. (Or skip the widget and continue on the next page.)
(Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade.)
URL: http://www.purplemath.com/modules/solveabs2.htm
© 2018 Purplemath. All right reserved. Web Design by