Absolute value equations with extraneous solutions

AI Thread Summary
The discussion centers on the confusion surrounding absolute value equations and the emergence of extraneous solutions. A specific case involving the equation |2x - 3| = x - 5 is highlighted, where 8/3 is incorrectly identified as a solution despite yielding contradictory results when substituted back into the equation. Participants agree that 8/3 is not a valid solution, emphasizing the importance of verifying solutions to avoid errors. The conversation also raises the question of how to identify extraneous solutions and stresses the necessity of checking all potential answers in absolute value equations. Ultimately, the consensus is that careful verification is crucial in solving these types of equations.
pempem
Messages
4
Reaction score
0
I'm a little confused about solving absolute value equations and why sometimes solutions don't seem to make sense.

Take a look at case (ii) on this website:
http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_AbsoluteValueEquations.xml

I understand the process of solving the equation, and I understand how they arrive at the two solutions. The only problem is, 8/3 is not actually a solution! You can see this by the simple fact that the left side of the equation has to be positive (since the entire left side is inside the absolute value "brackets"). Since 8/3 - 5 is a negative number, it can't be equal to the left side! When you plug in 8/3 you get 7/3 = - 7/3 which does not make sense.

The website claims 8/3 is a solution, but it certainly doesn't seem like it is. Can someone explain, in a mathematical sense, why this discrepancy comes about? Is there any way to know that an answer is extraneous or should one always check solutions to absolute value equations to make sure they are indeed solutions?

Thanks!
 
Mathematics news on Phys.org
You are completely correct that 8/3 is NOT a solution to |2x- 3|= x- 5.

On the right 2(8/3- 3= 16/3- 9/3= 7/3 while on the right 8/3- 5= 8/3- 15/3= -7/3.


Unfortunately, there are a number of "algebra" and "mathematics" sites like this one that are full of errors.
 
Good, I thought I was going crazy haha

In that case, what about the second part of my question: why does this discrepancy happen? Is there any way to know that an answer is extraneous or should one always check solutions to absolute value equations to make sure they are indeed solutions?

Thanks for your response!
 
According to Wolfram Alpha , x=-2 is also not a solution. This equation has no solutions .
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top