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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?
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April 20th, 2017, 03:54 PM   #2
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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

$(x+2)^2 = \dfrac 8 3$
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April 20th, 2017, 03:58 PM   #3
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$x^2+4x = -\dfrac{4}{3}$

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?
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April 20th, 2017, 03:59 PM   #4
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gasp I came in a while 4 minutes before Skeeter!
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April 20th, 2017, 04:16 PM   #5
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gasp I came in a while 4 minutes before Skeeter!
even a blind squirrel finds an acorn now & then ...
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April 20th, 2017, 04:28 PM   #6
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even a blind squirrel finds an acorn now & then ...
i'm gonna need some aloe
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April 20th, 2017, 04:45 PM   #7
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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.
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April 20th, 2017, 04:54 PM   #8
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Quote:
Originally Posted by romsek View Post
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 ()?
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April 20th, 2017, 04:57 PM   #9
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Quote:
Originally Posted by GIjoefan1976 View Post
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 $

$(x + 2)^2 = -\dfrac 4 3 + 2^2 = \dfrac 8 3$
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April 21st, 2017, 01:04 PM   #10
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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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