Good morning ☕️
Answer:
x ≥ 2
Step-by-step explanation:
our expression is defined when x ≥ 1 ,
because , when x ≥ 1
x - 1 ≥ 0
x + 2√(x - 1) ≥ 0
x - 2√(x - 1) ≥ 0
[tex]Calculations:\\\\(\sqrt{x+2\sqrt{x-1} } -\sqrt{x-2\sqrt{x-1} } )^{2} =2^2\\\\then\\\\(x+2\sqrt{x-1} ) +(x-2\sqrt{x-1} ) -2(\sqrt{x+2\sqrt{x-1} })(\sqrt{x-2\sqrt{x-1} } )=4\\\\then\\\\2x - 2\sqrt{(x+2\sqrt{x-1} )(x-2\sqrt{x-1} ) }=4 \\\\then\\\\2x -2\sqrt{(x-2)^2} = 4\\\\then\\\\2x -2|x - 2| =4\\\\then\\\\x - |x-2|=2\\\\now,\\\\case1 : (x\geq2)\\ x - |x-2|=2 \\means \\x-(x-2)=2\\means\\2=2\\\\case2: (1 \leq x \leq2)\\x - |x-2|=2 \\means \\x-(2-x)=2\\means\\2x-2=2\\means\\x=2[/tex]
According to case 1 and 2 the solution set for this equality is {x real number where x ≥ 2 }.
