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steve816
 one year ago
Simple algebra question help!
solve by completing the square.
4x^2+4x=3
steve816
 one year ago
Simple algebra question help! solve by completing the square. 4x^2+4x=3

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know what the first step is?

steve816
 one year ago
Best ResponseYou've already chosen the best response.0I believe we have to factor out the 4x^2+4x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you always want to set the equation to 0 first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0and do you know how to factor from there?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[4x^2+4x=3\]\[4(x^2+x)=3\]\[4\left(x^2+x+\color{red}{\frac{1}{4}}\right)=3+\color{red}{1} ~~\therefore \left(\frac{1}{\color{blue}{2}}\right)^2=\frac{1}{4}\]\[4\left(x+\frac{1}{2}\right)^2 = 4\]\[\boxed{4\left(x+\frac{1}{2}\right)^2 4 = 0}\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1I 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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