SOLUTION: Hi, this is the problem I was told to do: [SQRT(2x + 2)] - [SQRT(x - 3)] = 2 or √(2x+2) - √(x-3) = 2 I've been told to FOIL this problem, but I don't understand

Algebra ->  Radicals -> SOLUTION: Hi, this is the problem I was told to do: [SQRT(2x + 2)] - [SQRT(x - 3)] = 2 or √(2x+2) - √(x-3) = 2 I've been told to FOIL this problem, but I don't understand      Log On


   

Question 163137: Hi, this is the problem I was told to do:
[SQRT(2x + 2)] - [SQRT(x - 3)] = 2
or
√(2x+2) - √(x-3) = 2
I've been told to FOIL this problem, but I don't understand how to FOIL it if its subtracting and not multiplying. What I do know is that the answer is equal to 7.
Thanks in advance.
Found 2 solutions by jim_thompson5910, MathTherapy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let A=sqrt%282x%2B2%29 and B=sqrt%28x-3%29

So the equation sqrt%282x%2B2%29-+sqrt%28x-3%29=2 then becomes A-B=2

A-B=2 Start with the given equation


A=2%2BB Add "B" to both sides


A%5E2=%282%2BB%29%5E2 Square both sides. Remember, squaring undoes the square root (which is in both A and B)


A%5E2=4%2B4B%2BB%5E2 FOIL the right side


%28sqrt%282x%2B2%29%29%5E2=4%2B4%2Asqrt%28x-3%29%2B%28sqrt%28x-3%29%29%5E2 Plug in A=sqrt%282x%2B2%29 and B=sqrt%28x-3%29


2x%2B2=4%2B4%2Asqrt%28x-3%29%2Bx-3 Square sqrt%282x%2B2%29 to get 2x%2B2. Square sqrt%28x-3%29 to get x-3.


2x%2B2=1%2Bx%2B4%2Asqrt%28x-3%29 Combine like terms.


2x%2B2-1-x=4%2Asqrt%28x-3%29 Subtract 1 from both sides. Subtract x from both sides.


x%2B1=4%2Asqrt%28x-3%29 Combine like terms.


%28x%2B1%29%5E2=16%28x-3%29 Square both sides. This will eliminate the square root.


x%5E2%2B2x%2B1=16%28x-3%29 FOIL the left side


x%5E2%2B2x%2B1=16x-48 Distribute


x%5E2%2B2x%2B1-+16x%2B48=0 Get all terms to the left side


x%5E2-14x%2B49=0 Combine like terms


%28x-7%29%5E2=0 Factor the left side


x-7=0 Take the square root of both sides (to get rid of the square)


x=7 Add 7 to both sides.


So the solution is x=7

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, this is the problem I was told to do:

[SQRT(2x + 2)] - [SQRT(x - 3)] = 2
or
√(2x+2) - √(x-3) = 2

I've been told to FOIL this problem, but I don't understand how to FOIL it if its subtracting and not
multiplying. What I do know is that the answer is equal to 7.
Thanks in advance.
*****************
      sqrt%282x+%2B+2%29+-+sqrt%28x+-+3%29+=+2
This might be easier and less confusing if one of the radicals is moved to the right side. So, let's choose -+sqrt%28x+-+3%29 to do that with. 
                    sqrt%282x+%2B+2%29+=+2+%2B+sqrt%28x+-+3%29 ----- Adding  sqrt%28x+-+3%29 to both sides
               %28sqrt%282x+%2B+2%29%29%5E2+=+%282+%2B+sqrt%28x+-+3%29%29%5E2 -- Squaring both sides
               %28sqrt%282x+%2B+2%29%29%5E2+=+%282+%2B+sqrt%28x+-+3%29%29%282+%2B+sqrt%28x+-+3%29%29 
                        2x+%2B+2+=+4+%2B+2sqrt%28x+-+3%29+%2B+2sqrt%28x+-+3%29+%2B+%28sqrt%28x+-+3%29%29%5E2 ---- FOILing right-side
                        2x+%2B+2+=+4+%2B+4sqrt%28x+-+3%29+%2B+x+-+3
                        2x+%2B+2+=+1+%2B+4sqrt%28x+-+3%29+%2B+x
             2x+%2B+2+-+1+-+x+=+4sqrt%28x+-+3%29
                          x+%2B+1+=+4sqrt%28x+-+3%29
                      %28x+%2B+1%29%5E2+=+%284sqrt%28x+-+3%29%29%5E2 ---- Squaring both sides
                 x%5E2+%2B+2x+%2B+1+=+16%28x+-+3%29
                 x%5E2+%2B+2x+%2B+1+=+16x+-+48
x%5E2+%2B+2x+%2B+1+-+16x+%2B+48+=+0
            x%5E2+-+14x+%2B+49+=+0
                     %28x+-+7%29%5E2+=+0
                          x - 7 = 0___x = 7

             You can do the CHECK!!