Why is this the result of this simple equation with absolute value

[itsrayex]

This is the equation I am trying to solve:

The way I solved it it yelds three results, not two. Why so?
Here's how I solved it:

Thanks to anyone who will help

 

[pka]

This is the equation I am trying to solve:
View attachment 30894
The way I solved it it yelds three results, not two. Why so?
Here's how I solved it:
View attachment 30895
Thanks to anyone who will help

I wish that I could read your handwriting. But alas I cannot.
I will tell you that because the absolute value is always non-negative,
it must be the case that [imath]5-x\ge 0[/imath] or [imath]x\le 5[/imath].

[imath][/imath][imath][/imath]

 

[Deleted member 4993]

This is the equation I am trying to solve:
View attachment 30894
The way I solved it it yelds three results, not two. Why so?
Here's how I solved it:
View attachment 30895
Thanks to anyone who will help

These problems, with absolute value, can have extra solutions. Thus you will have to check whether some solutions need to be "discarded".

 

[Dr.Peterson]

This is the equation I am trying to solve:
View attachment 30894
The way I solved it it yelds three results, not two. Why so?
Here's how I solved it:
View attachment 30895
Thanks to anyone who will help

Your first case assumes that the expression inside the absolute value is non-negative, which requires that [imath]x\ge 2[/imath] or [imath]x\le 1[/imath]. Since 3 satisfies that, it is a valid solution.

Your second case assumes that the expression inside the absolute value is negative, which requires that [imath]1< x< 2[/imath]. Only the second solution shown is valid.

You can either keep track of these details as you work, or just check answers at the end.

 

[Harry_the_cat]

If you sub your solution \(\displaystyle x=4+\sqrt5\) into the original equation, the RHS will be negative BUT the LHS is non-negative (because it is an absolute value), so that solution is NOT valid.