sqrt(x+6+2 sqrt(x+5))+sqrt(x-1-sqrt(x+5))=4
x = 4
sqrt(4 + 6 + 2sqrt(4+5)) + sqrt(4-1-sqrt(4+5)) = 4
sqrt(10 + 2sqrt(9)) + sqrt(3 - sqrt(9)) = 4
sqrt(10 +- 6) + sqrt(3 +-3) = 4
The +- happens because, for example, the square roots of 9 are -3 and 3, not just 3.
Since we have two sets of +-, we have four possibilities:
#1)
sqrt(10 +- 6) + sqrt(3 +-3) = 4
sqrt(10+6) + sqrt(6) = 4
sqrt(16) + sqrt(6) = 4
+-4 + sqrt(6) = 4
sqrt(6) = 0 or 8
Neither of these is true, so this possibility can't happen.
.
#2)
sqrt(10 +- 6) + sqrt(3 +-3) = 4
sqrt(10+6) + sqrt(0) = 4
sqrt(16) = 4
+-4 = 4
True when we choose the positive square root of 16.
.
#3)
sqrt(10 +- 6) + sqrt(3 +-3) = 4
sqrt(10 - 6) + sqrt(3 + 3) = 4
sqrt(4) + sqrt(6) = 4
+-2 + sqrt(6) = 4
sqrt(6) = 2 or 6
Neither of these is true, so this possibility can't happen.
.
#4)
sqrt(10 +- 6) + sqrt(3 +-3) = 4
sqrt(10 - 6) + sqrt(3 - 3) = 4
sqrt(4) + sqrt(0) = 4
+-2 = 4
This is never true, so this possibility can't happen.
.
Answer:
sqrt(x+6+2 sqrt(x+5))+sqrt(x-1-sqrt(x+5))=4 with x = 4 can be true, but only if:
We take the positive square root of sqrt(x+5) (the left side).
We take the negative square root of sqrt(x+5) (the right side).
We take the positive square root of sqrt(x+6+2 sqrt(x+5)) (the outer sqrt).
