How do you solve #-2\sqrt { 2x } - 5= - 15#?

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Answer 1 · 2016-11-30T22:28:08

#x = 25/2#

First, isolate the #sqrt()# term:

#-2sqrt(2x) - 5 + 5 = -15 + 5#

#-2sqrt(2x) - 0 = -10#

#-2sqrt(2x) = -10#

#(-2sqrt(2x))/-2 = (-10)(-2)#

#sqrt(2x) = 5#

We can now square both sides and solve for #x#:

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

#2x = 25#

#(2x)/2 = 25/2#

#x = 25/2#