Solve for x

Here squaring both sides is not going to give any extra root.

2x + 2\sqrt{x^2 - 4x + 4} = 4

\Rightarrow x + |x-2| = 2 ___(1)

If x\leq 2 we can write the equation as x + 2-x = 2, So all x\leq 2 are solutions of equation (1).

But from the original equation we see x\geq 1.

x\epsilon [1,2]

Firstly i take that you wanted real solutions

solution is x≥1

Shubhodip right answer.

\sqrt{x-2 \sqrt{x-1}} = \sqrt{(x-1)-2 \sqrt{x-1}+1} = |\sqrt{x-1}-1|

Likewise

\sqrt{x+2 \sqrt{x-1}} = \sqrt{(x-1)-2 \sqrt{x-1}+1} = \sqrt{x-1}+1

Hence when x≥2, the expression equals 2\sqrt{x-1}

and when x≤2, it equals 2.

Hence [1,2] is the solution set

how about this:

lets try x=3

√(3+2√(3-1)) + √(3-2√(3-1))=(1+√2) +(1-√2)=2,clearly x=3 is a solution!

it is (1+√2) +(√2 - 1) and not what u wrote...

hope it helped

Since theprophet sir has shown that when x≥2, the expression is equal to 2√(x-1) , where √ is the positive square root , you must work with that expression....