Question

completing a square: 3a^2+18a+76=8

Answer 1

Hey tiff,
\(\Large\bf \color{#CC0033}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

Completing the Square huh? Mmmm let's see..

Answer 3

We would like our equation to be in this form,
\[\Large\rm \color{orangered}{x^2+bx}=stuff\]So we can complete the square on the x's.

Answer 4

Oops, a's instead of x's :)

Answer 5

We currently have:\[\Large\rm 3\color{orangered}{a^2}+18\color{orangered}{a}+76=8\]Let's subtract 76 to the other side,\[\Large\rm 3\color{orangered}{a^2}+18\color{orangered}{a}=8-76\]\[\Large\rm 3\color{orangered}{a^2}+18\color{orangered}{a}=-68\]We need to get that leading coefficient off of the a^2 term.
So we'll divide each side by 3,\[\Large\rm \color{orangered}{a^2}+\frac{18}{3}\color{orangered}{a}=-\frac{68}{3}\]

Answer 6

What'dyou think Tiff? :d
Following along?

Answer 7

\[\Large\rm \color{orangered}{a^2+6a}=-\frac{68}{3}\]

Answer 8

Yep!!

Answer 9

So now we've got it in that nice form that we were looking for!\[\Large\rm a^2+ba=stuff\]To complete the square, we take half of our b coefficient and square it,
\(\Large\rm \left(\dfrac{b}{2}\right)^2\)
This will be the value that we'll add to each side. It will complete our square for us.

Answer 10

So what is our b coefficient?
What value are we going to end up adding to each side?

What do you think tiff tiff? :O

Answer 11

9!

Answer 12

Ok great, our b coefficient is the 6, half of that is 3, squared is 9.\[\Large\rm \color{orangered}{a^2+6a}+9=-\frac{68}{3}+9\]

Answer 13

would it be (a+3)^2=-41/3?

Answer 14

Mmm yes very good! :)

And depending on what you are trying to accomplish.. you would do one two things next.
If you wanted to `solve for a`, you would take the square root of each side.
Or more commonly, we're trying to often put something into `vertex form`, so we would want to add our 41/3 to the other side.

Answer 15

i'm trying to find a

Answer 16

Or were we just trying to complete the square on a?
In that case I guess it wouldn't matter much where the 41/3 is,
but it's nice to leave it with the other stuff,\[\Large\rm (a+3)^2+\frac{41}{3}=0\]

Answer 17

If you're trying to `solve for a`, then we didn't want to go through the trouble of completing the square :)
There's a much quicker way to solve for a. Unless that's the approach your teacher wanted you to take.

Answer 18

can there be a square root on the denominator?

Answer 19

\[\Large\rm \sqrt{(a+3)^2}=\sqrt{-\frac{41}{3}}=\frac{\sqrt{-41}}{\sqrt3}\]Yes.

Answer 20

Wow! Okay, thank you sooooooo much!

Answer 21

Think you can handle it from there? :)
Remember how to deal with imaginary numbers and stuff?

Bahh I gotta go get ready for work >.<
No Problem \c:/
Good job.

Answer 22

Yep, i remember! thank you again!