Question #223480

(2+√y)^2-(2-√y)^2=8√y but in your solution it is equal to 4√y what's the reason.


1
Expert's answer
2021-08-05T13:50:22-0400

(2+y)2(2y)2=((2+y)+(2y))((2+y)(2y))=(2+2+yy)(22+y+y)=(4)(2y)=8yHence,(2+y)2(2y)2=8y4y(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}\\


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