SOLUTION: 3) For the equation x - 2 sqrt x = 0, perform the following: a) Solve for all values of x that satisfies the equation. Answer: Show work in this space. b) Graph th

Algebra ->  Algebra  -> Radicals -> SOLUTION: 3) For the equation x - 2 sqrt x = 0, perform the following: a) Solve for all values of x that satisfies the equation. Answer: Show work in this space. b) Graph th      Log On

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Question 36576: 3) For the equation x - 2 sqrt x = 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 y = 2 sqrt 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 rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
The basic strategy in solving a radical equation is to isolate the radical. x+-+2%2Asqrt%28x%29+=+0.

You can do this by adding %2B2%2A+sqrt%28x%29+ to each side:
x=+2%2A+sqrt%28x%29

Now, square both sides:
x%5E2+=+%282%2Asqrt%28x%29%29%5E2
x%5E2+=+4%2A+x

This is a quadratic equation, so set it equal to zero by subtracting 4x from each side:
x%5E2+-+4x=+0
x%28x-4%29+=+0
x=0 or x = 4

Check for extraneous solutions. Make sure these answers check!!
x=0
x=+2%2A+sqrt%28x%29
0=+2%2A+sqrt%280%29+
0=2%2A0 It checks!

x= 4
x=+2%2A+sqrt%28x%29+
4=+2%2A+sqrt%284%29+ It checks!

Now graph y+=+x and y+=+2%2Asqrt%28x%29, and find the points of intersection:

+graph+%28300%2C+300%2C+-6%2C6%2C+-6%2C6%2C+x%2C+2%2Asqrt%28x%29+%29
Notice how the graphs cross at x = 0 and at x = 4.

R^2 at SCC