sqrt 4 - x - sqrt x+6 = 2?

Question

I know I am supposed to move one of the radicals to the right with the 2 but I don't seem to be able to come up with a correct solution. Please help!

Answer 1

You do have to move one of the radicals over then square both sides, isolate the remaining radical, then square both sides again.

Might as well move the - sqrt x+6 over, getting

sqrt(4-x) = 2 + sqrt(x+6)

square both sides, remembering the right will be FOIL
4 - x = [2 + sqrt(x+6)][2 + sqrt(x+6)] = 4 + 4 sqrt(x+6) + (x+6)

4 - x = 10 + 4 sqrt(x+6) + x

move over the 10 and the x by adding their opposite

-2x - 6 = 4 sqrt(x+6)

square both sides again, FOIL on the left this time

4x^2 + 24x + 36 = 16 (x+6) = 16x + 96

move all over to get 0

4x^2 + 8x - 60 = 0; divide by 4 to get

x^2 + 2x - 15 = 0

Factor by FOIL to get two possible values of x; test both because usually one doesn't work (extraneous solution)

Answer 2

you meant sqr(4-x) - sqr(x+6) = 2 right?

square both sides
4-x + 2*sqr( (4-x)(x+6) + x+6 = 4

combine like terms
10 + 2 sqr( (4-x)(x+6) ) = 4

subtract 10

2 sqr ( (4-x)(x+6) ) = -6

square both sides
4 * (4-x)(x+6) = 36

foil
4 * (-x^2 - 2x + 24) = 36

distribute
-4x^2 - 8x + 96 = 36

-4x^2 - 8x + 60= 0

take out -4
-4(x^2 + 2x - 15) = 0

factor
-4(x+5)(x-3) = 0

x= -5 or 3

check: 3 is not solution. So -5 is the answer

Answer 3

4^2 -x- x^2+6= 2
16 -x-x^2 +6 = 2
16 -x-x^2 +4 = 0
20-x -x^2 =0
(4-x)(5+x) = 0
4-x = 0 or 5+x = 0
x = 4 x = -5

20 -(4)-(4^2) =0
20 - 4- 16= 0
0 = 0

20 -(-5) - 5^2 = 0
20 + 5 -25 = 0
0 = 0

Answer 4

huh? This is just basic math.
All X to one side all other stuff to the other.
-x - sqx = 2 -sq4 -6