steve816
  • steve816
Simple algebra question help! solve by completing the square. 4x^2+4x=3
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
do you know what the first step is?
steve816
  • steve816
I believe we have to factor out the 4x^2+4x
anonymous
  • anonymous
you always want to set the equation to 0 first

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anonymous
  • anonymous
you can do it that way, where you do (b/2)^2 , but I find it easier to just set it to 0. You can do it both ways though
anonymous
  • anonymous
so you could do it as : \[4x^2+4x=3\] \[(4/2)^2 = (2)^2 = 4\] add the four to both sides so it would look like: \[4x^2+4x+4=3+4\]
anonymous
  • anonymous
and do you know how to factor from there?
steve816
  • steve816
yes thanks so much!
anonymous
  • anonymous
You're welcome (:
Jhannybean
  • Jhannybean
\[4x^2+4x=3\]\[4(x^2+x)=3\]\[4\left(x^2+x+\color{red}{\frac{1}{4}}\right)=3+\color{red}{1} ~~\therefore \left(\frac{1}{\color{blue}{2}}\right)^2=\frac{1}{4}\]\[4\left(x+\frac{1}{2}\right)^2 = 4\]\[\boxed{4\left(x+\frac{1}{2}\right)^2 -4 = 0}\]
Jhannybean
  • Jhannybean
I completed the square `inside` the parenthesis `after` factoring out the common 4 from the LHS of the equation. In step 3, the `1` came from multiplying both sides of the equation by the common term, but in this case I had a 4 factored out of my parenthesis, and to keep it consistent, when i multiply `1/4` by `4`, i get a 1, which is what i add to the RHS of the equation. To simplify the square, just remember that it will be \(a\left(x\pm \dfrac{b}{2} \right)^2\pm c=0\)

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