(2+√y)^2-(2-√y)^2=8√y but in your solution it is equal to 4√y what's the reason.
(2+y)2−(2−y)2=((2+y)+(2−y))((2+y)−(2−y))=(2+2+y−y)(2−2+y+y)=(4)(2y)=8yHence,(2+y)2−(2−y)2=8y≠4y(2+\sqrt{y})^2-(2-\sqrt{y})^2\\ =((2+\sqrt{y})+(2-\sqrt{y}))((2+\sqrt{y})-(2-\sqrt{y}))\\ =(2+2+\sqrt{y}-\sqrt{y})(2-2+\sqrt{y}+\sqrt{y})\\ =(4)(2\sqrt{y})\\ =8\sqrt{y}\\ \hspace{0.0cm}\\ \textsf{Hence,}\\ (2+\sqrt{y})^2-(2-\sqrt{y})^2=8\sqrt{y}\not=4\sqrt{y}\\(2+y)2−(2−y)2=((2+y)+(2−y))((2+y)−(2−y))=(2+2+y−y)(2−2+y+y)=(4)(2y)=8yHence,(2+y)2−(2−y)2=8y=4y
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