Chapter 10 - Quadratic Equations - 10.1 - Quadratic Equations - Problem Set 10.1 - Page 442: 59

$a = 2\sqrt 7$

Plugging in values to the Pythagorean Theorem gives: $a^2+6^2 = 8^2$ Simplifying gives: $a^2 + 36 = 64$ Subtract 36 from both sides to obtain: $a^2 = 64-36 = 28$ To solve $a^2 = 28$, take the square root of both sides: $\sqrt a^2 = \sqrt28 = \sqrt4\times7$ Simplifying gives: $a=2\sqrt7$