SOLUTION: {{{sqrt(x) -sqrt(x-5)=1}}}

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Question 424269: sqrt%28x%29+-sqrt%28x-5%29=1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%29+-sqrt%28x-5%29=1
Here's a procedure you can use to solve square root equations like this:
  1. Isolate a square root that has the variable in its radicand. (The expression inside a radical is called a radicand.)
  2. Square both sides of the equation (carefully).
  3. If there is still a square root with the variable in its radicand, then repeat steps 1 and 2.
  4. At this point you should have no square roots with the variable in its radicand. Use techniques appropriate for the type of equation to solve it.
  5. Check your answer(s)! This is not optional. Whenever you square both sides of an equation, as is done at least once in step 2, extraneous solutions may be introduced. Extraneous solutions are solutions which fit the squared equation but do not fit the original equation. So answers must be checked. And any "solutions" that do not work are extraneous and must be rejected.
Let's see this in action:
1) Isolate a square root...
There are two square roots with variables in their radicands. We can isolate either one. I am going to isolate the first one because I can use addition to eliminate the second square root from the left side leaving us with an equation with no subtractions:
sqrt%28x%29+=+sqrt%28x-5%29%2B1

2) Square both sides.
%28sqrt%28x%29%29%5E2+=+%28sqrt%28x-5%29%2B1%29%5E2
The left side is simple to square. To square the right side correctly (since exponents do not distribute!), we must use FOIL or the %28a%2Bb%29%5E2+=+a%5E2+%2B+2ab+%2B+b%5E2 pattern. Both methods will give us the same result. I prefer using the pattern:
x+=+%28sqrt%28x-5%29%29%5E2+%2B+2%28sqrt%28x-5%29%29%281%29+%2B+%281%29%5E2
which simplifies as follows:
x+=+x-5+%2B+2sqrt%28x-5%29+%2B+1
x+=+x-4+%2B+2sqrt%28x-5%29+
Note how squaring both sides did not eliminate both square roots!

3) If there are still square roots...
There is still a square root with a variable in its radicand. So back to step 1.

1) Isolate a square root...
This time there is just one square root. Subtracting x and adding 4 we get:
4+=+2sqrt%28x-5%29+
The 2 in front of the square root is not a problem where it is. The right side will still square easily. But if it bothers you we can divide both sides by 2:
2+=+sqrt%28x-5%29

2) Square both sides:
%282%29%5E2+=+%28sqrt%28x-5%29%29%5E2
This time both sides are simple to square:
4 = x-5

3) If there is still a square root ...
There are no more square roots. On to step 4.

4) Solve the equation.
This is a very simple equation to solve. Just add 5 to each side:
9 = x

5) Check your answer(s).
Use the original equation to check:
sqrt%28x%29+-sqrt%28x-5%29=1
Checking x = 9:
sqrt%289%29+-sqrt%289-5%29=1
sqrt%289%29+-sqrt%284%29=1
3 - 2 = 1
1 = 1 Check!!

So the only solution to your equation is x = 9.