April 8th, 2017, 11:19 AM  #11 
Senior Member Joined: Jan 2015 From: London Posts: 102 Thanks: 2  
April 8th, 2017, 11:33 AM  #12  
Math Team Joined: Jul 2011 From: Texas Posts: 2,579 Thanks: 1275  Quote:
$[(x+3)^29]2 = (x+3)^2 + 9  2 = (x+3)^2 + 7 \ne (x+3)^2 + 11$ (2) because I prefer putting everything inside $[bracket]$. if you want to leave the +2 outside the $[bracket]$ ... $2  6x  x^2 = x^2  6x + 2$ $(x^2 + 6x) + 2$ $(x^2 + 6x + 9  9) + 2$ $\left[(x^2 + 6x + 9)  9\right] + 2$ $\left[(x+3)^2  9\right] + 2$ $(x+3)^2 + 9 + 2$ $(x+3)^2 + 11$  
April 8th, 2017, 12:28 PM  #13  
Senior Member Joined: Jan 2015 From: London Posts: 102 Thanks: 2  Quote:
Thanks for all the help  

Tags 
completing, square 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Completing The Square Help  Trainpancake  Algebra  3  October 2nd, 2013 02:11 PM 
completing the square  milly2012  Algebra  2  April 28th, 2012 08:44 AM 
Completing the square  bearsfan092  Algebra  3  January 24th, 2010 06:09 PM 
Completing the square  PistolPete  Algebra  15  May 20th, 2009 06:44 PM 
Completing the square  SteveThePirate  Algebra  4  May 17th, 2009 12:44 AM 