SOLUTION: For the equation x - 2SQRT of X=0 , perform the following: a) Solve for all values of x that satisfies the equation. b) Graph the functions y = x and y=2 SQRT of X on

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Question 77062: For the equation x - 2SQRT of X=0 , perform the following:
a) Solve for all values of x that satisfies the equation.



b) Graph the functions y = x and y=2 SQRT of X on the same graph (by plotting points if necessary). Show the points of intersection of these two graphs.
Graph:

Points of intersection:

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

For the equation x+-+2%2Asqrt%28x%29=0 , 
perform the following:
a) Solve for all values of x that 
satisfies the equation. 

x+-+2%2Asqrt%28x%29=0

Isolate the radical term by adding 2%2Asqrt%28x%29 to both sides

x+=+2%2Asqrt%28x%29 

Square both sides

x%5E2+=+%282%2Asqrt%28x%29%29%5E2

x%5E2+=+4x

Get 0 on the right

x%5E2+-+4x+=+0

Factor x out on the left

x%28x-4%29=0

Use the zero-factor property
and set each of the factors
x and x-4 equal to 0.

x+=+0 gives answer as 0
x-4=0 gives another answer as 4

We must check these in the original

x+-+2%2Asqrt%28x%29=0
0+-+2%2Asqrt%280%29=0
0+-+0+=+0
0+=+0
So 0 checks

x+-+2%2Asqrt%28x%29=0
4+-+2%2Asqrt%284%29=0
4+-+2%2A2+=+0
4-4=0
(((0=0)))
So 4 checks too.

b) Graph the functions y = x and 
y=2 SQRT of X on the same graph (by plotting 
points if necessary). Show the points of 
intersection of these two graphs.

The idea here is to take the equation
x+=+2%2Asqrt%28x%29 and set y = to each side.
Then graph the two equations. Then the two 
values of y should be equal when the two sides 
of x+=+2%2Asqrt%28x%29 are equal, and that
will be where the graphs intersect.

Some points on y=x are 
(-2,-2), (-1,-1) (2,2) (5,5)

Some points on y=2%2Asqrt%28x%29 are 
(.5,1.4), (1,2), (3,3.5), (5,4.5)

+graph%28+300%2C+300%2C+-5%2C+5%2C+-5%2C+5%2C0%2C++x%2C+2%2Asqrt%28x%29%29+

So we see that the blue curve and the green 
line intersect at (0,0) and (4,4) and the 
x-values are 0 and 4 at those points, which 
correspond to the algebraic solution for x 
above.

Edwin