SOLUTION: how to solve the square root of x+4= the square root of x plus the square root of 2

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Question 714216: how to solve the square root of x+4= the square root of x plus the square root of 2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%2B4%29=sqrt%28x%29%2Bsqrt%282%29
First a bit of vocabulary. "Radicand" is the name for the expression inside a radical.

Here's a procedure for solving equations where the variable is in the radicand of a square root:
  1. Isolate a square root that has the variable in its radicand.
  2. Square both sides of the equation. Squaring the isolated square root will be easy. Squaring the other side can easily be done incorrectly! This is where many mistakes occur so be careful!
  3. If there are still any square roots with the variable in its radicand, then repeat steps 1-3.
  4. At this point there should no longer be any square roots with the variable in its radicand. Use appropriate techniques for whatever kind of the equation the equation now is.
  5. Check your solution(s). This is not optional! Whenever you square both sides of an equation, like step 2, "extraneous" solutions may occur. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. Extraneous solutions may occur any time both sides of an equation is raised to an even power (like squaring). They are not an indication of an error. You just have to check for them and, if found, reject them.
Let's see this in action:
1. Isolate a square root.
The square root on the left side is already isolated.

2. Square both sides.
%28sqrt%28x%2B4%29%29%5E2=%28sqrt%28x%29%2Bsqrt%282%29%29%5E2
Squaring the left side is simple. Squaring the right side correctly requires either FOIL or use of the %28a%2Bb%29%5E2+=+a%5E2%2B2ab%2Bb%5E2 pattern. Personally I prefer using the pattern:
x%2B4=%28sqrt%28x%29%29%5E2%2B2%28sqrt%28x%29%29%28sqrt%282%29%29%2B%28sqrt%282%29%29%5E2
Simplifying...
+x%2B4+=+x+%2B+2sqrt%282x%29+%2B+2

3. If there are still square roots...
There is still a square root with the variable in its radicand. So we repeat theses steps.
1. Isolate a square root.
Subtracting x and 2 from both sides we get:
2+=+2sqrt%282x%29
The square root is sufficiently isolated. Mathematically the 2 in front does not cause a problem. But if it bothers you, go ahead and divide both sides by 2.

2. Square both sides.
%282%29%5E2+=+%282sqrt%282x%29%29%5E2
which simplifies as follows...
4+=+2%5E2%28sqrt%282x%29%29%5E2
4+=+4%2A2x
4+=+8x

3. If there is still a square root...
There are no square roots remaining so on to step 4.

4. Solve the equation.
Just divide both sides by 8:
4%2F8+=+x
which simplifies to
1%2F2+=+x

5. Check the solution(s).
Use the original equation to check:
sqrt%28x%2B4%29=sqrt%28x%29%2Bsqrt%282%29
Checking x+=+1%2F2:
sqrt%28%281%2F2%29%2B4%29=sqrt%28%281%2F2%29%29%2Bsqrt%282%29
So far so good. (Some extraneous solutions make a radicand negative which must be rejected.) Simplifying...
sqrt%284.5%29=sqrt%280.5%29%2Bsqrt%282%29
For a complete, perfect check we would go through the steps above to eliminate the square roots. I'm going to see how close the decimal approximations are:
2.1213203435596425732025330863145 = 0.70710678118654752440084436210485 + 1.4142135623730950488016887242097
2.1213203435596425732025330863145 = 2.1213203435596425732025330863145
The decimal approximations seem to confirm that our solution is correct (and not an extraneous solution).