How do you solve #sqrt(x-5) = 2sqrt6#?

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Answer 1 · 2016-05-25T21:59:42

#x=29#

Square both sides.

#(sqrt(x-5))^2=(2sqrt6)^2#

On the left side, the square root and the exponent of #2# undo one another, leaving just #x-5#.

On the right side, to square #2sqrt6#, we see that #(2sqrt6)^2=2^2*(sqrt6)^2=4*6=24#.

#x-5=24#

#x=29#

Answer 2 · 2016-05-25T21:59:57

#x = 29#

square both sides
#x - 5 = (2sqrt6)^2 -> x - 5 = 2^2 xx 6 -> x - 5 = 24#
add 5 to both sides
#x = 29#