Chapter 6 - Radical Functions and Rational Exponents - 6-5 Solving Square Root and Other Radical Equations - Practice and Problem-Solving Exercises - Page 395: 13

$5$

Square both sides ($\sqrt {2x-1}$)$^2$ = (3)$^2$ $2x-1$ = $9$ Add $1$ to each side $2x-1+1=9+1$ $2x=10$ Divide by $2$ to botho sides: $\dfrac{2x}{2}=\dfrac{10}{2}$ $x=5$ Substitute into original problem: $\sqrt {2(5)-1} = 3$ $\sqrt {10-1} = 3$ $\sqrt {9} = 3$ $3=3$