SOLVING ABSOLUTE VALUE EQUATIONS WITH EXTRANEOUS SOLUTIONS

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Extraneous solutions are values that we get when solving equations that aren't really solutions to the equation.

To find extraneous solution for absolute value function, we follow the steps given below.

(i) Decompose the absolute function in the form of

Absolute sign on one side

Absolute sign on both sides

Note :

In the branches, first two are the same and last two are the same.

(ii) Apply the values that we have received into the original question and check which value satisfies the equation of not.

The values that doesn't satisfy the equation is known as extraneous solution.

Solve each equation. Check for extraneous solution.

Problem 1 :

|x – 1| = 5x + 10

Solution :

|x – 1| = 5x + 10

x - 1 = 5x + 10 (or) x - 1 = -(5x + 10)

x – 1 = 5x + 10

x - 5x = 10 + 1

-4x = 11

x = -11/4

x – 1 = -5x – 10

x + 5x = -10 + 1

6x = -9

x = -3/2

Checking extraneous solution :

If x = -11/4

|x – 1| = 5x + 10

|-11/4 – 1| = 5(-11/4) + 10

|-15/4| = -15/4

15/4 ≠ -15/4

If x = -3/2

|x – 1| = 5x + 10

|(-3/2) – 1| = 5(-3/2) + 10

|-5/2| = (-15 + 20)/2

5/2 = 5/2

So, -3/2 is a solution and -11/4 is an extraneous solution.

Problem 2 :

|2z – 3| = 4z - 1

Solution :

|2z – 3| = 4z - 1

2z - 3 = 4z - 1

2z – 4z = -1 + 3

-2z = 2

z = -1

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

2z – 3 = -4z + 1

2z + 4z = 1 + 3

6z = 4 ==> z = 2/3

Checking extraneous solution :

If z = -1

|2z – 3| = 4z – 1

|2(-1) – 3| = 4(-1) – 1

|-2 – 3| = -4 – 1

|-5| = -5

5 ≠ -5

If z = 2/3

|2z – 3| = 4z - 1

|2(2/3) – 3| = 4(2/3) - 1

|4/3 – 3| = 8/3 - 1

|-5/3| = 5/3

5/3 = 5/3

So, -1 is a extraneous solution.

Problem 3 :

|3x + 5| = 5x + 2

Solution :

Decomposing into two branches.

3x + 5 = 5x + 2

3x – 5x = 2 – 5

-2x = -3

x = 3/2

3x + 5 = -5x – 2

3x + 5x = -2 – 5

8x = -7

x = -7/8

Checking extraneous solution :

If x = 3/2

|3x + 5| = 5x + 2

|3(3/2) + 5| = 5(3/2) + 2

|19/2| = 19/2

19/2 = 19/2

If x = -7/8

|3x + 5| = 5x + 2

|3(-7/8) + 5| = 5(-7/8) + 2

|-21/8 + 5| = -35/8 + 2

19/8 ≠ -19/8

So, -7/8 is the extraneous solution.

Problem 4 :

|2y – 4| = 12

Solution :

Decomposing into two branches.

2y – 4 = 12

2y = 12 + 4

2y = 16

y = 8

2y – 4 = -12

2y = -12 + 4

2y = -8

y = -4

Checking extraneous solution :

If y = 8

|2y – 4| = 12

|2(8) – 4| = 12

|16 – 4| = 12

12 = 12

y = 8 is a solution.

If y = -4

|2(-4) – 4| = 12

|-8 – 4| = 12

|-12| = 12

12 = 12

y = 4 is a solution.

So, there is no extraneous solution.

Problem 5 :

3|4w – 1| - 5 = 10

Solution:

3|4w – 1| - 5 = 10

Adding 5 on both sides.

3|4w – 1| - 5 + 5 = 10 + 5

3|4w – 1| = 15

Dividing 3 on both sides.

|4w – 1| = 5

4w – 1 = 5

4w = 5 + 1

4w = 6

w = 3/2

4w – 1 = -5

4w = -5 + 1

4w = -4

w = -1

Check for extraneous solution :

If w = 3/2

3|4w – 1| - 5 = 10

3|4(3/2)– 1| - 5 = 10

3|5| - 5 = 10

3|5| = 15

5 = 5

If w = -1

3|4(-1) – 1| - 5 = 10

3|-4 – 1| - 5 = 10

3|-5| = 10 + 5

3|-5| = 15

5 = 5

So, there is no extraneous solution.

Problem 6 :

|2x + 5| = 3x + 4

Solution :

|2x + 5| = 3x + 4

2x + 5 = 3x + 4

5 – 4 = 3x – 2x

1 = x

2x + 5 = -3x – 4

5 + 4 = -3x – 2x

9 = -5x

-9/5 = x

Check for extraneous solution :

If x = 1

|2x + 5| = 3x + 4

|2(1) + 5| = 3(1) + 4

7 = 7

If x = -9/5

|2x + 5| = 3x + 4

|2(-9/5) + 5| = 3(-9/5) + 4

|-18/5 + 5| = -27/5 + 4

|7/5| = -7/5

7/5 ≠ -7/5

So, -9/5 is a extraneous solution.

Problem 7 :

You are driving on a highway and are about 250 miles from your state’s border. You set your cruise control at 60 miles per hour and plan to turn it off within 30 miles of the border on either side. Find the minimum and maximum numbers of hours you will have cruise control on.

Solution :

One way to solve is to write an absolute value equation that models the number x of hours you will have cruise control on. You know that the distance you travel will be within 30 miles of 250 miles.

|60x - 250| = 30

60x - 250 = 30

60x = 30 + 250

60x = 280

x = 280/60

x = 4.6

60x - 250 = -30

60x = -30 + 250

60x = 220

x = 220/60

x = 3.6

you will travel at least 3.6 hours and at most 4.6 hours with cruise control on.

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SOLVING ABSOLUTE VALUE EQUATIONS WITH EXTRANEOUS SOLUTIONS

Absolute sign on one side

Absolute sign on both sides

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