Chapter 9 - Solving Quadratic Equations - 9.5 - Solving Quadratic Equations Using the Quadratic Formula - Exercises - Page 521: 14

$x=\frac{1}{3}$

Comparing $9x^{2}-6x+1=0$ with $ax^{2}+bx+c=0$, we get $a=9$, $b=-6$ and $c=1$. The quadratic formula is $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Substituting the values, we get $x=\frac{-(-6) \pm \sqrt {(-6)^{2}-4(9)(1)}}{(2)(9)}=\frac{6}{18}=\frac{1}{3}$