$\textbf{Solve for $\mathbf{x::}$}$\\\\ \mathbf{\sqrt{x+2.\sqrt{x-1}}+\sqrt{x-2.\sqrt{x-1}}=2}$
21Here 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]
341\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
1how 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!
1Since 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....
