Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 32

$n=5$

$Given,$ $n=\sqrt (4n+5)$ Squaring both sides,we get: $n^{2}=4n+5$ $n^{2}-4n-5=0$ $(n-5)(n+1)=0$ $n=5,n=-1$ We need to check by the putting the obtained solutions in the original equation. $n=5$ $L.H.S=5$ $R.H.S=\sqrt (4X5+5)=\sqrt 25=5$ $L.H.S=R.H.S$,hence it is a valid solution $x=-1$ $L.H.S=-1$ $R.H.S=\sqrt (4X-1+5)=1$ $L.H.S$ does not equal $R.H.S$,hence it is an extraneous solution.