How am I solve this?

[shahar]

Can I say the two expressions are equal 0, because square expression cannot be negative.
And from there we assume that the expression are equal to zero.

And:
when the expression 2x-1 = 0 the second expression doesn't equal to zero, and from there x = 1/2 and 2 + 1/2 = 21/2 that isn't zero.
Q.E.D. (quod erat demonstrandum)

 

[Deleted member 4993]

View attachment 27104
Can I say the two expressions are equal 0, because square expression cannot be negative.
And from there we assume that the expression are equal to zero.

And:
when the expression 2x-1 = 0 the second expression doesn't equal to zero, and from there x = 1/2 and 2 + 1/2 = 21/2 that isn't zero.
Q.E.D. (quod erat demonstrandum)

Your thought process is correct. As posted, that equation does not have a solution.

 

[Dr.Peterson]

View attachment 27104
Can I say the two expressions are equal 0, because square expression cannot be negative.
And from there we assume that the expression are equal to zero.

And:
when the expression 2x-1 = 0 the second expression doesn't equal to zero, and from there x = 1/2 and 2 + 1/2 = 21/2 that isn't zero.
Q.E.D. (quod erat demonstrandum)

Yes.

Now, if you didn't notice that, and solved by the usual method (moving one radical and squaring both sides), you would get the "solution" x = 3, which is extraneous -- it is really the solution to \(\sqrt{2x-1}-\sqrt{x+2}=0\), not to the equation you were given.