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
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.
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