How do you solve #8=sqrt( 2x-3)+3#?

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Answer 1 · 2016-01-31T08:24:29

#sqrt(2x-3)+3=8#

#rarrsqrt(2x-3)=8-3#

#rarrsqrt(2x-3)=5#

Square both sides:

#rarr(sqrt(2x-3))^2=5^2#

#rarr2x-3=25#

#rarr2x=5+3#

#rarr2x=8#

#rarrx=8/2=4#