SOLUTION: how do i simplify the square root of 6-x minus the square root of x-2 equals 2

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Question 279491: how do i simplify the square root of 6-x minus the square root of x-2 equals 2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%286-x%29+-+sqrt%28x-2%29+=+2
Nothing can be simplified but we can solve the equation for x. To solve equations where the variable is inside a square root:
  1. Isolate a square root.
  2. Square both sides of the equation (remembering that exponents do not distribute).
  3. If there is still a square root, repeat steps #1 and #2.
  4. Solve the equation (which should not have any square roots remaining).
  5. Check your answers! This is not just a good idea. It is important. At step #2 we squared both sides of an equation. Squaring both sides of an equation is not wrong. But it can introduce what are called extraneous solutions. Extraneous solutions are solutions which fit the squared equation but do not fit the original (pre-squared) equation. We must check for these extraneous solutions and reject them.

Let's see how this works on your equation.
1. Isolate a square root.
By adding the second square root to each side, the first one will be by itself (i.e. isolated) and I'll get rid of a "-":
sqrt%286-x%29+=+sqrt%28x-2%29+%2B+2

2. Square both sides:
%28sqrt%286-x%29%29%5E2+=+%28sqrt%28x-2%29+%2B+2%29%5E2
The left side is easy to simplify. On the right side, to square it properly, we should use FOIL or the pattern %28a%2Bb%29%5E2+=+a%5E2+%2B2ab+%2B+b%5E2 (or other proper technique for multiplying two-term expressions):
6-x+=+%28sqrt%28x-2%29%29%5E2+%2B+2%28sqrt%28x-2%29%29%282%29+%2B+%282%29%5E2
6-x+=+x-2+%2B+4sqrt%28x-2%29+%2B+4
6-x+=+x+%2B+2+%2B+4sqrt%28x-2%29

3. If there is still a square root repeat steps #1 and #2. We still have s square root so...
Isolate a square root. There is only one square root so that's the one we need to isolate. Subtracting x and 2 from each side:
4-2x+=+4sqrt%28x-2%29
Dividing by 4:
%284-2x%29%2F4+=+sqrt%28x-2%29
Square both sides:
%28%284-2x%29%2F4%29%5E2+=+%28sqrt%28x-2%29%29%5E2
%284%5E2+-+2%2A%284%29%282x%29+%2B+%282x%29%5E2%29%2F4%5E2+=+x-2
%2816+-+16x+%2B+4x%5E2%29%2F16+=+x-2
There are no more square roots so we can move to step 4.

4. Solve.
We'll get rid of the fraction first by multiplying both sides by 16:
16+-+16x+%2B+4x%5E2+=+16x-32
This is a quadratic equation so we'll get one side equal to zero (by subtracting 16x from and adding 32 to each side):
4x%5E2+-32x+%2B48+=+0
Now we factor:
4%28x%5E2+-8x+%2B12%29+=+0
4(x-2)(x-6) = 0
By the Zero Product Property we know that this product is zero only if one of the factors is zero. 4 cannot be zero but the other two factors can:
x-2=0 or x-6=0
Solving these we get:
x = 2 or x = 6

5. Check the solutions.
Always use the original equation to check.
sqrt%286-x%29+-+sqrt%28x-2%29+=+2
Checking x = 2:
sqrt%286-%282%29%29+-+sqrt%28%282%29-2%29+=+2
sqrt%284%29+-+sqrt%280%29+=+2
2+-+0+=+2
2+=+2 Check!

Checking x = 6:
sqrt%286-%286%29%29+-+sqrt%28%286%29-2%29+=+2
sqrt%280%29+-+sqrt%284%29+=+2
0+-+2+=+2
-2+=+2 Does not check! (This is an extraneous solution which must be rejected.)

So the only solution to your equation is x = 2.