Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 28

$a=\frac{-1}{2}$ is extraneous

$Given,$ $2a=\sqrt (4a+3)$ Checking for $a=\frac{3}{2}$: $L.H.S=2X\frac{3}{2}=3$ $R.H.S=\sqrt (4X\frac{3}{2}+3)=\sqrt (6+3)=\sqrt 9=3$ $L.H.S=R.H.S$,hence not extraneous Checking for $a=\frac{-1}{2}$: $L.H.S=2X\frac{-1}{2}=-1$ $R.H.S=\sqrt (4X\frac{-1}{2}+3)=\sqrt (-2+3)=\sqrt 1=1$ $L.H.S$ does not equal $R.H.S$,hence extraneous