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

$z=3/4$

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