Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 623: 13

$a=16$

$Given,$ $\sqrt (3a+1)=7 $ Squaring,we get: $3a+1=49$ $3a+1-1=49-1$ $3a=48$ $a=48\div3=16$ We need to check by putting the obtained value in the original equation : $L.H.S=\sqrt (3X16+1)=\sqrt 49=7=R.H.S$ Hence,$a=16$ is a valid solution