Question 72601
<pre><font size = 5><b>I am kind of stuck on this problem...
maybe someone can assist me! any help 
would be greatly appreciated! 
                      _
For the equation x - <font face = "symbol">Ö</font>x = 0, perform 
the following:
a) Solve for all values of x that satisfies 
the equation.
       _
  x - <font face = "symbol">Ö</font>x = 0

Isolate the radical term:
            _
       x = <font face = "symbol">Ö</font>x 

Square both sides:

             _
    (x)² = (<font face = "symbol">Ö</font>x)² 

      x² = x

Get 0 on the right:

  x² - x = 0

Factor left side

x(x - 1) = 0

Set each factor = 0:

x = 0,  x - 1 = 0
            x = 1

Now we must check these answers in
the original equation:
       _
  x - <font face = "symbol">Ö</font>x = 0

Checking x = 0
       _
  0 - <font face = "symbol">Ö</font>0 = 0  
   0 - 0 = 0
       0 = 0

Checking x = 1    
       _
  x - <font face = "symbol">Ö</font>x = 0
       _
  1 - <font face = "symbol">Ö</font>1 = 0
   1 - 1 = 0
       0 = 0

b) Graph_the functions y = x and
   y = <font face = "symbol">Ö</font>x on the same graph 
(by plotting points if necessary). 

{{{ graph( 300, 300, -2, 2, -2, 2, 0, x, sqrt(x)) }}} 

Show the points of intersection of these two graphs. 

The green line is the graph_of y = x and the blue 
curve is the graph of y = <font face = "symbol">Ö</font>x.
They intersect at the two points (0,0) and (1,1).

c) How does the graph relate to part a?

If you wanted to find the points where the green line
crosses the blue curve, you would solve the system:

y = x_
y = <font face = "symbol">Ö</font>x

If you subtract the equations you get
         _
0 = x - <font face = "symbol">Ö</font>x

 or
     _
x - <font face = "symbol">Ö</font>x = 0

which is the original equation
Then you would solve that as above and get
x = 0, x = 1. Then you would substitute
these in one of the equations of the
system above and get y = 0, and y = 1
respectively. So the two points where
the blue curve and and green line intersect
are at (0, 0) and (1, 1).

Edwin</pre>