Question: Solve: \dfrac{x -9}{\sqrt{x}+3} = 1

Solution:
Given,

\dfrac{x -9}{\sqrt{x}+3} = 1

or, \dfrac{(\sqrt{x})^2 -(3)^2}{\sqrt{x}+3} = 1

or, \dfrac{(\sqrt{x} + 3)(\sqrt{x} - 3) }{\sqrt{x} + 3} = 1

or, \sqrt{x} - 3 = 1

or, \sqrt{x} = 3 +1

or, \sqrt{x} = 4

[ Squaring both sides ]

or, (\sqrt{x})^2 = 4^2

\therefore x = 16
= Answer