How do you solve using the completing the square method 9x^2 - 6x + 1=0?

1 Answer
Jun 9, 2018

x=-1/3

Explanation:

9x^2-6x+1=0

First, divide every term by 9

x^2-6/9x+1/9=0

Move your 1/9 to the RHS

x^2-6/9x=-1/9

Simplify 6/9

x^2-2/3x=-1/9

Now, your general form of a quadratic equation is ax^2+bx+c=0
To complete the square, your equation looks something like this ax^2+bx+(b/2)^2=-c

What this means is that you divide your coefficient of x by 2 and then square your entire number

x^2-2/3x +((2/3)/2)^2=-1/9+((2/3)/2)^2

x^2-2/3x+(1/3)^2=-1/9+1/9

x^2-2/3x+(1/3)^2=0

(x+1/3)^2=0

x=-1/3