Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 31

$x=3$

$Given,$ $x=\sqrt (2x+3)$ Squaring both sides,we get: $x^{2}=2x+3$ $x^{2}-2x-3=0$ $(x-3)(x+1)=0$ $x=3,x=-1$ We need to check by the putting the obtained solutions in the original equation. $x=3$ $L.H.S=3$ $R.H.S=\sqrt (2X3+3)=\sqrt 9=3$ $L.H.S=R.H.S$,hence it is a valid solution $x=-1$ $L.H.S=-1$ $R.H.S=\sqrt (2X-1+3)=1$ $L.H.S$ does not equal $R.H.S$,hence it is an extraneous solution.