Precalculus help struggling with some absolute value questions. Thanks!

Question

Solve the following                                   |  =  absolute value sign

 

2 | 2x²+4x-6|  + 4 = 2

 

 

| 3x -4| = x-2              

 

Thanks!

 

       ps, I realize these questions are pretty simple but I have missed classes lately due to illness and am unaware of what to do exactly, thanks again                     

Answer 1

If I told you that |x| = 6, you'd tell me that x could be one of two things: 6 or -6. That's the basic idea behind solving absolute value problems: Whatever quantity is inside the absolute value (x) is either the thing on the other side of the equals sign (6) or its negative (-6).

 

So, you'll first want to get the absolute value by itself on one side. Then, you'll solve for possible answers. Finally, you'll want to check your solutions, in case any are extraneous (sometimes, solutions don't actually work).

 

By the way, remember that if you get the absolute value by itself and have a negative on the other side, there are no solutions. |2x - 3| = -10, for example. There is nothing I can put in for x that will give me -10, because absolute values make everything positive.

 

 

 

(1)

 

2 |2x² + 4x - 6| + 4 = 2

2 |2x² + 4x - 6| = -2

|2x² + 4x - 6| = -1

 

That's impossible, since there's a negative on the right-hand side. So, there are no solutions.

 

 

 

Let's do this problem as well, just so you see how to solve it:

 

-2 |2x² + 4x - 6| + 4 = 2

-2 |2x² + 4x - 6| = -2

|2x² + 4x - 6| = 1

 

Now, the quantity inside the absolute value either equals 1 or -1.

 

 

IF IT EQUALS 1:

 

2x² + 4x - 6 = 1

2x² + 4x - 7 = 0

 

Use the Quadratic Formula to solve for x. You'll get x = 1.12 or x = -3.12.

 

 

IF IT EQUALS -1:

 

2x² + 4x - 6 = -1

2x² + 4x - 5 = 0

 

Use the Quadratic Formula again to get x = 0.87 or x = -2.87.

 

 

So, you've got four solutions for x: 1.12, -3.12, 0.87, and -2.87. It's a good idea to check those solutions to make sure they all work. Plug each of those values in for x and solve.

 

 

 

(2)

 

|3x - 4| = x - 2

 

The quantity inside the absolute value equals either x - 2 or -(x - 2).

 

 

IF IT EQUALS x - 2:

 

3x - 4 = x - 2

2x - 4 = -2

2x = 2

x = 1

 

 

IF IT EQUALS -(x - 2):

 

3x - 4 = -(x - 2)

3x - 4 = -x + 2

4x - 4 = 2

4x = 6

x = 3/2

 

 

So, x = 1 or x = 3/2. But think about those answers one more time. Let's plug in x = 1:

 

|3(1) - 4| = 1 - 2

|3 - 4| = -1

|-1| = -1

1 = -1

 

That's not actually true. This is an extraneous solution. Let's check x = 3/2:

 

|3(3/2) - 4| = 3/2 - 2

|9/2 - 4| = -1/2

|1/2| = -1/2

1/2 = -1/2

 

Again, that's an extraneous solution. So, there are actually no solutions. This is why it's important to check whatever answers you get! This kind of thing often happens with absolute values.

 

 

 

Let's do this problem as well, just so you see how to solve it:

 

|3x - 4| = x + 2

 

 

IF IT EQUALS x + 2:

 

3x - 4 = x + 2

2x - 4 = 2

2x = 6

x = 3

 

 

IF IT EQUALS -(x + 2):

 

3x - 4 = -(x + 2)

3x - 4 = -x - 2

4x - 4 = -2

4x = 2

x = 1/2

 

 

So, x = 3 or x = 1/2. Plug those in to see that they both work.

Answer 2

Well, math is well known as one of those subjects that has prerequisites (that is, new lessons build on information contained in old lessons), so if you miss something ,for whatever reason, you are less capable of doing future work. 

p.s., I’m working on computer software that will allow adults – using the principles of andragogy, adult learning, not pedagogy, student learning -- that would allow working people to “fill in” blanks or weak spots in their formal education in ways that are much nicer than “going back to school” (the remedial approach).  For example, a sales person that hated English class in school and often makes embarrassing grammatical errors or the (for real) wonderful air force mechanic that I tutored who flunked high school math, but now wanted to take the AFOQT test (and become a pilot).

So, absolute values (expressions enclosed within “|” vertical bars are always positive.  Thus |3| < 4, and |-3) < 4.   If you compute the value of the expression, then drop the sign, you’re there !

The laws of algebra that you’ve learned so far still apply as long as you have an equation.  It will change when you learn about “<” and “>.”

So, for this problem:

     2 * |2x² + 4x| + 4 = 2

You may rearrange to be:

       |2x² – 4x| = 4

    Which is true for 2x² -4x = -4    and for 2x² -4x = 4       so, list the two possible values of x.

AMD, be sure to check your results.

     No real value of x is possible.

 

Then,   |3x – 4| = x – 2

This is true when 3x-4 = x-2   and it is true for    -(3x-4) = x -2

 Again, there are two answers.

AND, be sure to check your results !

      Is  |3*(1) – 4| =  (1)  - 2

              |-1| = -1              NO!  This is NOT a solution.

      Is  |3*(3/2) – 4| = (3/2)  - 2

           | 9/2 – 4 | = - 1/2

            | - 1/2|   =  -1/2               AGAIN, this is NOT a solution.

Thus, there are also no real solutions to this problem either.

This should illustrate why checking your answer should be half the grade for your solution !

Answer 3

2  | 2x² + 4x - 6 | + 4 = 2

 

 

Subtract 4 on both sides of equation.

 

2  | 2x² + 4x - 6 | = -2

 

 

Divide both sides of equation by 2 to isolate the absolute value term.

 

| 2x² + 4x - 6 | = -1

 

 

 

The absolute value of any number is always positive.  Therefore, there are no solutions to this problem.

 

 

 

 

| 3x - 4 | = x - 2

 

 

The absolute value of a negative number is a positive value.  The absolute value of a positive number is also a positive value.  Knowing this, we can break up the equation into two parts.

 

3x - 4 = x - 2                       and                        3x - 4 = -(x - 2)             

                                                                       

                                                                        3x - 4 = -x + 2

 

2x - 4 = - 2                          and                        4x - 4 = 2

 

2x = 2                                  and                       4x = 6

 

  x = 1                                  and                         x = 6/4

                                  

                                                                          x = 3/2

 

 

We have two solutions.

 

x = 1

x = 3/2

 

Answer 4

The first equation has no solution. To see this, first get the absolute value by itself by subtracting 4 and dividing by 2:

 

2|2x²+4x-6|=2-4=-2

 

|2x²+4x-6|=-1

 

This can never be satisfied since the absolute value of a number is always at least 0. Therefore, there is no solution.

 

For the second equation to be satisfied, either

 

-(3x-4)=x-2 needs to be true

or

(3x-4)=x-2

 

In order to find *potential* solutions, we can solve these two equations for x. I emphasize that they are potential solutions and will still need to be checked once we find them.

 

The first equation:

 

-(3x-4)=x-2

 

Distribute the negative:

 

-3x+4=x-2

 

Get the x terms on one side:

 

-4x+4=-2

Subtract 4:

 

-4x=-6

 

x=3/2

 

The other equation is

 

3x-4=x-2.

 

Solving this one like the previous one gives

 

2x=2

x=1

 

Our two candidates for solutions are x=3/2 and x=1. We need to check both of them by plugging them into the original equation. It turns out that both of them fail:

 

|3(3/2)-4|=|9/2-4|=|1/2|=1/2,

but

(3/2)-1=-1/2, so the equality is not satisfied. 

 

For the other candidate, x=1,

 

|3(1)-4|=|3-4|=|-1|=1, but

(1)-2=-1,

 

so the equation is still not satisfied, and the second equation also has no solution.

 

I hope this helps.

 

EDIT:

 

It has been interesting to see a variety of tutors pipe in and give different perspectives on this problem. In case anyone is interested, I took the liberty of making a diagram of the problem that shows where the extraneous solutions come from and uploaded it with some explanatory text here: https://www.wyzant.com/resources/files/356195/an_absolute_value_problem_with_no_solution_v_2

Feel free to have a look. I hope this has been helpful and instructive. Also, thanks to David W. for feedback on this writeup. I have edited it to reflect his suggestions.

Answer 5

2 | 2x² + 4x - 6 | + 4 = 2

2 | 2x² + 4x - 6 | = -2

   | 2x² + 4x - 6 | = -1             The absolute value of an expression is always positive. No solution.

 

Equations containing absolute values must be solved twice: once assuming the expression is positive, once assuming the expression is negative.

 

1. 3x - 4 > 0

 

3x - 4 = x - 2

2x - 4 = -2

2x = 2

x = 1

 

Check this solution

 

| 3(1) - 4 | = 1 - 2

| 3 - 4 | = -1               Absolute value is always positive.  

 

2. 3x - 4 < 0

 

-(3x - 4) = x - 2

-3x + 4 = x - 2

-4x + 4 = -2

-4x = -6

x = 3/2

 

Check this solution

| 3(3/2) - 4 | = 3/2 - 2

| 3(3/2) - 4 | = -1/2        Absolute value is positive.

 

No solutions.