Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 623: 7

$x=4$

$Given,$ $\sqrt x + 3=5$ $\sqrt x + 3-3=5-3$ $\sqrt x=2$ Squaring on both sides, $x=4$ We need to check by putting the obtained value in the original equation: $L.H.S=\sqrt 4+3=2+3=5=R.H.S$ Hence,$x=4$ is a valid solution