April 20th, 2017, 03:49 PM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 354 Thanks: 10  completing the square
HI, I am on what I believe is the third step of solving this problem so I am about to divide the 4x by 2 and then put it over 2 then divide it by 2 which would be 1 then one squared gets us one then we add 1 to each side? Would that be a good start? Thanks. Last edited by skipjack; April 21st, 2017 at 01:59 PM. 
April 20th, 2017, 03:54 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717 
no. You want to get the $x^2$ term to have a coefficient of $1$ $x^2 + 4x = \dfrac 4 3$ is a good place to start we take half the coefficient of the $x$ term $\dfrac 4 2 = 2$ and add and subtract it's square to get $x^2 + 4x + 4  4 = \dfrac 4 3$ now group the 3 left terms $(x+2)^2  4 = \dfrac 4 3$ and now bring the constant term from the left over to the right $(x+2)^2 = \dfrac 8 3$ 
April 20th, 2017, 03:58 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,628 Thanks: 1308 
$x^2+4x = \dfrac{4}{3}$ halve the coefficient of the linear term, square, and add the result to both sides ... $x^2 + 4x + 4 = \dfrac{4}{3} + 4$ left side is a square trinomial (that's why they call it "complete the square") ... $(x+2)^2 = \dfrac{8}{3}$ can you finish? 
April 20th, 2017, 03:59 PM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717 
gasp I came in a while 4 minutes before Skeeter! 
April 20th, 2017, 04:16 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,628 Thanks: 1308  
April 20th, 2017, 04:28 PM  #6 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717  i'm gonna need some aloe Last edited by romsek; April 20th, 2017 at 04:30 PM. 
April 20th, 2017, 04:45 PM  #7 
Senior Member Joined: Feb 2016 From: seattle Posts: 354 Thanks: 10 
hmmm, sorry if you all said this but so I would divide 4x by 2 then square it and I would end up with 4 ? as the video I am watching says to divide 4 by 2 then put it over 2 so that it looks like this (2/2)^2 so this is why I was thinking 2 divided by 2 equals 1 and then 1 squared is 1?
Last edited by skipjack; April 21st, 2017 at 02:03 PM. 
April 20th, 2017, 04:54 PM  #8  
Senior Member Joined: Feb 2016 From: seattle Posts: 354 Thanks: 10  Quote:
ahh if I am hopefully understanding it I take the 4 out of the 4x and put it over 2 and then divide it by 2 and then only square after I have divided it? which gets us 4. if so then my problem was I was dividing the 4 before I had put it in the ()?  
April 20th, 2017, 04:57 PM  #9  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717  Quote:
let's go back to zeroth principles $(x + a)^2 = x^2 + 2ax + a^2$ so if you find yourself with $x^2 + b x = c$ $x^2 + b x + \left(\dfrac b 2 \right)^2 = c + \left(\dfrac b 2 \right)^2$ $\left(x + \dfrac b 2\right)^2 = c + \left(\dfrac b 2 \right)^2$ in this case with $b=4,~c = \dfrac 4 3 $ $(x + 2)^2 = \dfrac 4 3 + 2^2 = \dfrac 8 3$  
April 21st, 2017, 01:04 PM  #10 
Senior Member Joined: Feb 2016 From: seattle Posts: 354 Thanks: 10 
So if I have one like this... 7xx^2/2=0 I multiply by 2 so that I get 14xx^2=0, could I then divide everything by negative 1 so I then get x^ 2+14=0? Last edited by skipjack; April 21st, 2017 at 02:05 PM. 

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