Chapter 9.5, Problem 14E

To determine

To solve the equation using Quadratic formula.

Answer to Problem 14E

  $x=\frac{1}{3}$

Explanation of Solution

Given:

Equation:

  $9x^2-6x+1=0$

Calculation:

On comparing the equation with $a{x}^2+bx+c$

  $a=9,b=-6,c=1$

Finding the value of $x$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values,

  $x=\frac{-(-6)\pm \sqrt{{(-6)}^2-4(9)(1)}}{2(9)}$

Opening brackets,

  $x=\frac{6\pm \sqrt{36-36}}{18}$

Solving root,

  $x=\frac{6\pm 0}{18}$

So,

  $x=\frac{6}{18}$

Therefore,

  $x=\frac{1}{3}$