Question: Solve: $\sqrt{x -7} = \sqrt{x} - 1$

Solution:
Given,

$\sqrt{x -7} = \sqrt{x} - 1$

[ squaring both sides ]

$or, (\sqrt{x -7})^2 = (\sqrt{x} - 1)^2$

$or, x -7 = (\sqrt{x})^2 - 2× \sqrt{x}×1 + 1^2$

$or, x - 7 = x -2\sqrt{x} +1$

$or, x - x = - 2\sqrt{x} + 1 +7$

$or, 0 + 2\sqrt{x} = 8$

$or, \sqrt{x} = \dfrac{8}{2}$

$or, \sqrt{x} = 4$

[ Squaring both sides ]

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

$\therefore x = 16$
= Answer