My Math Forum completing the square

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April 20th, 2017, 03:49 PM   #1
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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.
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Last edited by skipjack; April 21st, 2017 at 01:59 PM.

 April 20th, 2017, 03:54 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,206 Thanks: 614 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: 2,516 Thanks: 1239 $x^2+4x = -\dfrac{4}{3}$ 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: CA Posts: 1,206 Thanks: 614 gasp I came in a while 4 minutes before Skeeter!
April 20th, 2017, 04:16 PM   #5
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Quote:
 Originally Posted by romsek gasp I came in a while 4 minutes before Skeeter!
even a blind squirrel finds an acorn now & then ...

April 20th, 2017, 04:28 PM   #6
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Quote:
 Originally Posted by skeeter even a blind squirrel finds an acorn now & then ...
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: 343 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
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Quote:
 Originally Posted by romsek 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 $(x+2)^2 = \dfrac 8 3$

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
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Quote:
 Originally Posted by GIjoefan1976 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 but 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?

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: 343 Thanks: 10 So if I have one like this... 7x-x^2/2=0 I multiply by 2 so that I get 14x-x^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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