Expert: Paul Klarreich - 12/10/2010

Question
(3x-1)^2=7
I have gotten this far but can figure out da resh
(3x-1)(3x-1)=7
9x^2-3x-3x+1=7
9x^2-6x+1=7
              -7=-7
9x^2-6x-6=0
3(3x^2-2x-2)=0

Answer
Questioner:   Marilyn
Country:  United States
Category:  Advanced Math
Private:  No
 
Subject:  Quadratics
Question:  (3x-1)^2=7
I have gotten this far but can figure out da resh

>> you mean "the rest", don't you?


(3x-1)(3x-1)=7
9x^2-3x-3x+1=7
9x^2-6x+1=7
             -7=-7
9x^2-6x-6=0
3(3x^2-2x-2)=0
...................
If you don't mind my saying so, this is a terrible way to do it.  Since 7 is not a perfect square, this equation would not be factorable, so there is no point in trying.  Instead, use the method of 'Completing the Square', which is already almost done for you:

(3x-1)^2 = 7

3x - 1 = +- sqrt(7)

3x = 1 +- sqrt(7)

    1 +- sqrt(7)
x = -------------
         3

That is it.  I think you will find that applying the quadratic formula to your last line:

3(3x^2-2x-2)=0

3x^2 - 2x - 2 = 0

with a = 3,  b = -2, c = -2

will give that result.

Learn to use (and love??) completing the square.