Answer
$x=7$ or $x=3$
Work Step by Step
$\sqrt {2x - 5} - \sqrt {x - 3} = 1$
Isolate radical on one side by adding $\sqrt {x-3}$ on both sides.
$\sqrt {2x - 5} - \sqrt {x - 3} + \sqrt {x-3}= 1 +\sqrt {x-3}$
Square both sides.
$(\sqrt {2x - 5})^2 = (1 +\sqrt {x-3})^2$
$2x - 5 = 1^2 +2\sqrt {x-3}+(\sqrt {x-3})^2 $
Combine like terms.
$2x - 5 = 1 +2\sqrt {x-3}+x-3 $
$x - 3 = 2\sqrt {x-3}$
Square both sides.
$(x - 3)^2 = (2\sqrt {x-3})^2$
$x^2 - 6x+9 = 4(x-3)$
$x^2 - 6x+9 = 4x-12$
Combine like terms
$x^2 - 6x+9 = 4x-12$
$x^2 - 10x+21 = 0$
Use the quadratic formula to solve for $x$:
$a=1$, $b=-10$, $c=21$
$x = \frac{-b±\sqrt{b^2-4ac}}{2a}$
$x = \frac{-(-10)±\sqrt{(-10)^2-(4⋅1⋅21)}}{2⋅1}$
$x = \frac{10±\sqrt{100-84}}{2}$
$x = \frac{10±\sqrt{16}}{2}$
$x = \frac{10±4}{2}$
$x = \frac{10+4}{2}$ or $x = \frac{10-4}{2}$
$x=7$ or $x=3$
