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

Question

6594 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-09T08:39:11

#x=-1/3#

#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#