How do you simplify #(sqrt5)^2#?

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Answer 1 · 2016-09-12T19:03:42

#sqrt(5^2) = sqrt25 = 5" " (sqrt5)^2 = 5#

Adding and subtracting are inverse operations:

#15 color(blue)(+ 11)color(red)( -11) = 15" " 15 rarr 15#

Multiplication and division are inverse operations:

#23 color(blue)(xx 11) color(red)(div 11) = 23 " " 23 rarr 23#

Square and square root are inverse operations:

#color(blue)(sqrt(17)^color(red)(2)) = (color(blue)(sqrt17))^color(red)(2) = 17 " " 17 rarr17#

#sqrt(x^2) = (sqrtx)^2 = x" " x rarr x#

Inverse operations cancel each other out,

"One does, the other undoes."

The original number ends up the same.

#sqrt(5^2) = sqrt25 = 5" " (sqrt5)^2 = 5" " 5 rarr5#