## Intermediate Algebra for College Students (7th Edition)

$x=7$ or $x=3$
$\sqrt {2x - 5} - \sqrt {x - 3} = 1$ 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 $x=3$