Algebra 1: Common Core (15th Edition)

$b=1$
$2b=\sqrt {b+3}$ $(2b)^2=(\sqrt {b+3})^2$ $4b^2=b+3$ $4b^2-b-3=0$ $4b^2+3b-4b-3=0$ $b(4b+3)-(4b+3)=0$ $(b-1)(4b+3)=0$ $b-1=0$ or $4b+3=0$ $b=1$ or $b=-\frac{3}{4}$ Check for extraneous solutions: $2*1=\sqrt {1+3}=2$ $2*\frac{-3}{4}=\frac{-3}{2},\sqrt {\frac{-3}{4}+3}=3/2$ (eliminate this solution).