Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 623: 14

$b=3$

$Given,$ $\sqrt (10b+6)=6 $ Squaring,we get: $10b+6=36$ $10b+6-6=36-6$ $10b=30$ $b=30\div10=3$ We need to check by putting the obtained value in the original equation : $L.H.S=\sqrt(10X3+6)=\sqrt 36=6=R.H.S$ Hence,$b=3$ is a valid solution