How do you solve #sqrt(4z+1)=3+sqrt(4z-2)#?

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Answer 1 · 2015-04-14T15:10:03

#z=3/4#

Solution

#sqrt(4z+1)=3+sqrt(4z-2)#

Squaring both sides

#(sqrt(4z+1))^2=(3+sqrt(4z-2))^2#

#=>(4z+1)=(3)^2+(sqrt(4z-2))^2+2(3)sqrt(4z-2)#

#=>4z+1=9+4z-2+6sqrt(4z-2)#

#=>0=-4z-1+9+4z-2+6sqrt(4z-2)#

#=>0=-cancel(4z)-1+9+cancel(4z)-2+6sqrt(4z-2)#

#=>0=6+6sqrt(4z-2)#

#=>-6=6sqrt(4z-2)#

#=>-6/6=sqrt(4z-2)#

#=>-1=sqrt(4z-2)#

Again squaring both sides

#(-1)^2=(sqrt(4z-2))^2#

#=>1=4z-2#

#=>1+2=4z#

#=>3=4z#

#=>3/4=z#

Or

#z=3/4#