SOLUTION: For the equation x-2sqrtx=0 , perform the following: a) Solve for all values of x that satisfies the equation. Answer: Show work in this space. b) Graph the func

Algebra ->  Graphs -> SOLUTION: For the equation x-2sqrtx=0 , perform the following: a) Solve for all values of x that satisfies the equation. Answer: Show work in this space. b) Graph the func      Log On


   

Question 77193: For the equation x-2sqrtx=0 , perform the following:
a) Solve for all values of x that satisfies the equation.
Answer:
Show work in this space.




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

Points of intersection:

Found 2 solutions by stanbon, bucky:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
For the equation x-2sqrtx=0 , perform the following:
a) Solve for all values of x that satisfies the equation.
Answer:
Show work in this space.
2sqrtx=x
Square both sides after dividing both sides by 2 to get:
x=(x/2)^2
x=x^2/4
x^2-4x=0
x(x-4)=0
x=0 or x=4
============
b) Graph the functions y = x and on the same graph (by plotting points if necessary). Show the points of intersection of these two graphs.
Graph:
Points of intersection:
You did not post both of the functions.
The one you did post is a line thru (0,0), (1,1)
============
Cheers,
Stan H.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
First part of the problem:
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Given:
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x+-+2%2Asqrt%28x%29+=+0
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Find all the values of x that satisfy this equation.
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The method involves getting the radical term on one side of the equation and the other term
on the other side. We can do this by adding 2%2Asqrt%28x%29 to both sides and the equation
then becomes:
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x+=+2%2Asqrt%28x%29
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Next square both sides. Note that when you do that, the right side squares as follows:
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%282%2Asqrt%28x%29%29%5E2+=+%282%5E2%29%2A%28sqrt%28x%29%29%5E2+=+4%2Ax
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and the equation therefore squares to:
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x%5E2+=+4x
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Subtract 4x from both sides and the equation is then:
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x%5E2+-+4x+=+0
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Factor the common "x" from both terms on the left side to get:
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x%2A%28x+-+4%29+=+0
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Note that this equation will be true if either of the factors is zero. Therefore,
one at a time we can set the factors equal to zero to solve the equation. First:
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x+=+0
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That is one answer. Then:
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x-4+=+0
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Solve by adding 4 to both sides to get:
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x+=+4
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So the answer to the first part of this problem is that the equation given in the problem
will be satisfied by two values of x ... x = 0 and x = 4.
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For the second part of the problem the two graphs should look like this:
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graph%28600%2C400%2C-1%2C20%2C-5%2C+20%2C+x%2C+x+-+2%2Asqrt%28x%29%29
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Note that the original equation given in the problem does not permit x to be less than
zero because for real values you are not permitted to take the square root of a negative
number.
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The red line is the graph of y+=+x and the green line is the graph of y+=+x+-+2%2Asqrt%28x%29
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Note that the green line is on the x-axis at the two values of x that we found previously,
x = 0 and x = +4. Also note that the only place the two equations intersect is at the
origin. You can also show this is true mathematically by solving the equation set:
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y+=+x and y+=+x+-+2%2Asqrt%28x%29
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Solve by substitution. The first equation tells us that y equals x so we can substitute
x for y in the second equation. With this substitution, the second equation becomes:
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x+=+x+-+2%2Asqrt%28x%29
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Subtract x from both sides and the equation reduces to:
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0+=+-2%2Asqrt%28x%29
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Square both sides and you get:
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+0+=+4%2Ax
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and divide both sides by 4 to find that:
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x+=+0
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Then you can solve for the corresponding value of y by returning to the equation y+=+x
and plugging in 0 for x to find that y also = 0. The only common point on the two graphs
is (0,0) ... the origin as is shown on the graphs themselves.
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Hope this helps you to understand the problem and how to go about getting the solution
to it.
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