## Precalculus (6th Edition) Blitzer

$5-\sqrt 2$
Since $\sqrt 2 - 5$ is negative, we must negate it to find the distance from 0 and, therefore, the absolute value: $-(\sqrt 2 - 5) = 5-\sqrt 2$ . Therefore, $|\sqrt 2 - 5| = 5-\sqrt 2$.