How do you solve #sqrt(2x + 10 )- 6 = 2#?

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Answer 1 · 2016-02-14T07:21:10

#color(green)(x=27#

#color(blue)(sqrt(2x+10)-6=2#

Add #6# both sides:

#rarrcolor(blue)(sqrt(2x+10)=2+6#

#rarrcolor(blue)(sqrt(2x+10)=8#

Square both sides to get rid of the radical sign:

#rarrcolor(blue)((sqrt(2x+10))^2=8^2#

#rarrcolor(blue)(2x+10=64#

Subtract #10# both sides:

#rarrcolor(blue)(2x+10-10=64-10#

#rarrcolor(blue)2x=54#

Divide both sides by #2#:

#rarrcolor(blue)((2x)/2=54/2#

#rarrcolor(blue)((cancel2x)/cancel2=27#

#rArrcolor(green)(x=27#