How do you solve #sqrt( x + 7) = 2 sqrtx#?

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Answer 1 · 2017-06-29T10:01:01

Get rid of the square roots.

#(sqrt(x+7))^2= (2sqrtx)^2#

#x+7=4x#

#3x=7#

#x=7/3#

Answer 2 · 2017-06-29T10:03:07

The answer is #7/3#.

#sqrt (x+7)# = 2#sqrtx#

#(sqrt (x+7))^2# = #(2 sqrtx)^2#

#x + 7# = #4x#

#7# = #4x - x#

#7 = 3x#

#x = 7/3#