Questions on Algebra: Square root, cubic root, N-th root answered by real tutors!

To solve the problem correctly, we should reveal a context accurately.


The domain, i.e. the area where this expression is defined, is the set of points (x,y)
in the 1st quadrant or in the 3rd quadrant.  It means that the domain is the union of both quadrants.


In other words x >= 0, y >= 0 are allowed under the square root; or x <= 0, y <= 0 simultaneously.
But x and y of different signs are not allowed.


Second note is that when square root is considered as a function or as an expression,
by a commonly accepted agreement, it is considered as non-negative value or expression.


    Again, it is the common agreement for square root, when it is considered 
    as an expression of variables, that its output values are non-negative.



Next, the first move is to write   = .    (1)


But this expression is only half-right.  On the other half it is WRONG.



It is wrong in 3rd quadrant.  Numerical values under the square root are non-negative there. 
It is good, since we can extract the square root then.


But the values of the simplified expression (1) will be negative in 3rd quadrant, 
which is not consistent with the agreement mentioned above.


Therefore, we should write (1) in MODIFIED form

     =  = ,   (2)


placing x under the absolute value sign.


In this form (2),  it is consistent with (1) in the first quadrant and provides non-negative output
in the third quadrant.


ANSWER.  }  is    in the 1st and 3rd quadrants.

It's a little longer this way, but maybe you can understand it better
this way:



Write what's under the square root as a product of the simplest factors
possible:



Now group as many pairs of like factors as possible:

 

Now realize that when a pair of like factors are multiplied,
that is that factor squared. So write every pair as a square

 

Now we take the square root of every one of those squares,
and since the square root undoes the square we have all
those outside the square root, and only the x and the y
that didn't pair up have to be left inside the square root:



or



Edwin

To prove 

Let x and y be non-negative real numbers. 

lemma: square roots of non-negative real numbers are unique.

Proof:
For contradiction, suppose for positive real number " a ", p and q 
are two different non-negative square roots of " a ".  Then,

 =>  =>  
since , p-q=0, so p=q.  So we have reached a contradiction 
namely, that p and q are NOT different.

We show that the squares of  and  are equal.





Therefore 

and by the lemma, 

PROVED.

Edwin

Suppose a = -2

  Not equal!

But   

  Equal!

The definition of absolute value is 

When the radical  is used, it is always either positive
(or zero, in the case ), but  is never negative!

Although we often say "SQUARE ROOT OF X" when we see , we
really should say "NON-NEGATIVE SQUARE ROOT OF X".

[It's like in geometry, we often say "line" when we really should say
 "line segment".]

Edwin