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