SOLUTION: How do you redauce out the radicals in sqrt((x^2)+((y-2)^2)+(z^2)) + sqrt((x^2)+((y+2)^2)+(z^2)) = 8

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Question 616785: How do you redauce out the radicals in
sqrt((x^2)+((y-2)^2)+(z^2)) + sqrt((x^2)+((y+2)^2)+(z^2)) = 8
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!

Here's a general procedure for eliminating square roots from an equation:
  1. Isolate a square root.
  2. Square both sides. Be careful with this step. Squaring the isolated square root should be easy. But squaring the other side of the equation can be easy to get wrong.
  3. If there is still a square root, then repeat steps 1-3.
Let's see this in action:
1. Isolate a square root.
The equation has two square roots and neither one is by itself. It does not matter which square root you isolate. I usually isolate the square root with the simpler radicand (expression inside), Your radicands are about the same so I'm going to just pick one at random. Subtracting the second square root from each side we get:


2. Square both sides.

Squaring the left side is easy. Squaring the right correctly (Exponents do not distribute!!) requires that we use FOIL or the %28a-b%29%5E2+-+a%5E2-2ab%2Bb%5E2 pattern. I like using the pattern:



3. Repeat if there's still a square root.
1. Isolate a square root.
This time we only have one square root. So there's no choice to make. Subtracting every term of the right side, except the square root term, we get:

The next step is to square both sides. So the simpler we can make this the easier the next step will be. Because I can see even more will cancel, I'm going to multiply out the %28y-2%29%5E2 and %28y%2B2%29%5E2

-8y+-+64+=+-16sqrt%28%28x%5E2%29%2B%28%28y%2B2%29%5E2%29%2B%28z%5E2%29%29
And dividing both sides by -8 makes things even simpler:
y+%2B+8+=+2sqrt%28%28x%5E2%29%2B%28%28y%2B2%29%5E2%29%2B%28z%5E2%29%29
The square root is sufficiently isolated. The 2 is not a problem. If it bothers you, you can divide both sides by 2 before the next step...

2. Square both sides.
%28y+%2B+8%29%5E2+=+%282sqrt%28%28x%5E2%29%2B%28%28y%2B2%29%5E2%29%2B%28z%5E2%29%29%29%5E2

y%5E2+%2B+16y+%2B+64+=+4%2A%28%28x%5E2%29%2B%28%28y%2B2%29%5E2%29%2B%28z%5E2%29%29

3. Repeat if any square roots remain.

Rather than stop at the equation above, I think you should probably simplify and put the equation in standard form:
y%5E2+%2B+16y+%2B+64+=+4x%5E2%2B4%28y%2B2%29%5E2%2B4z%5E2
y%5E2+%2B+16y+%2B+64+=+4x%5E2%2B4%28y%5E2%2B4y%2B4%29%2B4z%5E2
y%5E2+%2B+16y+%2B+64+=+4x%5E2%2B4y%5E2%2B16y%2B16%2B4z%5E2
Subtracting the entire left side (because we want a zero on one side):
0=+4x%5E2%2B3y%5E2%2B4z%5E2%2B%28-48%29