How do you solve?

Question

√x + 1 - √x + 6 + 1 = 0

The x+1 is in a square root and x+6 is in a square root.

Answer 1

√(x + 1) - √(x + 6) + 1 = 0

1.) Isolate one of the items under the square root:
√(x + 1) + 1 = √(x + 6)

2.) Square both sides:
(√(x + 1) + 1)^2 = (√(x + 6))^2
(x+1) + 2√(x + 1) + 1 = x + 6

3.) Combine like terms.
(x+1) + 2√(x + 1) + 1 = x + 6
2√(x + 1) = x + 6 - (x+1) - 1
2√(x + 1) = 4
√(x + 1) = 2

4.) Square both side and simplify:
(√(x + 1))^2 = 2^2
(x + 1) = 4
x = 3 (solution!)

Check:
√(x + 1) - √(x + 6) + 1 = 0
√4 - √9 + 1 = 0
2 - 3 + 1 = 0
0 = 0 (check!)

Answer 2

This can be simplified as
sq root of (x+1) +1 = sq root of (x+6)
Therefore square of LHS + square of LHS
ie =x+1+[2 multiplied by sq root of (x+1) multiplied by 1]+1 = x+6
ie : sq root of (x+1) = [x+6-x-1-1] divided by 2
ie : sq root of (x+1) = 2
square both sides
ie : x+1=4
Therefore x=4-1
ie : X=3

Answer 3

If you can ask this question,then surely you know the answer.

Answer 4

wow, that is hard, noone knows that