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

1. anonymous

do you know what the first step is?

2. steve816

I believe we have to factor out the 4x^2+4x

3. anonymous

you always want to set the equation to 0 first

4. 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

5. 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: $4x^2+4x+4=3+4$

6. anonymous

and do you know how to factor from there?

7. steve816

yes thanks so much!

8. anonymous

You're welcome (:

9. anonymous

$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}$

10. anonymous

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 $$a\left(x\pm \dfrac{b}{2} \right)^2\pm c=0$$