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Mathematics 14 Online
OpenStudy (anonymous):

Create a quadratic equation and demonstrate how it would be solved by graphing, factoring, the quadratic formula, and by completing the square.

OpenStudy (anonymous):

Well can you create one for us?

OpenStudy (anonymous):

First off do you know what a quadratic equation looks like?

OpenStudy (anonymous):

I wouldn't know where to start...

OpenStudy (anonymous):

not right off the bat but I have been studying it.

OpenStudy (anonymous):

can you help me on the first to , I have an Idea on squaring but the other two not really.

OpenStudy (anonymous):

\[ax^2 + bx + c = 0\]

OpenStudy (anonymous):

okay.

OpenStudy (anonymous):

That's how a general quadratic equation looks like. Where a b and c are any values

OpenStudy (anonymous):

a usually equals 1 to make it easier to solve

OpenStudy (anonymous):

so I just have to fill thos in with #'s??

OpenStudy (anonymous):

Well like I said a is 1 to make it easier to solve. Then b and c have some connection to each other in order to use the complete square method

OpenStudy (anonymous):

so like 2 and 4?

OpenStudy (anonymous):

Well it's more complicated than that

OpenStudy (anonymous):

For example (x + 2) ( x+2)

OpenStudy (anonymous):

give you x^2 + 2x + 2x + 4= x^2 +4x+4

OpenStudy (anonymous):

yah, but that's after factoring right?

OpenStudy (anonymous):

in this case b=4 and c=4

OpenStudy (anonymous):

Yeah

OpenStudy (anonymous):

oh okay.

OpenStudy (anonymous):

But I'm talking about completing the square.

OpenStudy (anonymous):

yah, I got the hang of it now. But what about the quadratic factoring.

OpenStudy (anonymous):

It's hard for me to explain that part but try other example like (x+2)(x+1)

OpenStudy (anonymous):

To see what b and c equal

OpenStudy (anonymous):

uh-huh.

OpenStudy (anonymous):

(x+2)(x+1)= x^2 + 2x + x + 2= x^2 + 3x + 2

OpenStudy (anonymous):

in that case b=3 and c=2

OpenStudy (anonymous):

are we still talking about completing the square?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

okay.

OpenStudy (anonymous):

Lets say a=1 b=4 c=4 what's the general equation look like?

OpenStudy (anonymous):

it's going to be this equation ax^2 + bx + c = 0? or is this for somthing else?

OpenStudy (anonymous):

Yeah exactly like that

OpenStudy (anonymous):

so all these examples that you just did can be pluged in for that?

OpenStudy (anonymous):

and that's jusr for completing the aquare or general form now?

OpenStudy (anonymous):

No you can solve using quadratic formula for example \[\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array} \]

OpenStudy (anonymous):

You plug in a b and c to the left and solve for x

OpenStudy (anonymous):

ah, okay.

OpenStudy (anonymous):

\[\pm \] this sign means you have to do it two times. Solve it once for + and another one for - so you should get two answers.

OpenStudy (anonymous):

for x

OpenStudy (anonymous):

For the example that I gave you where a=1 b=2 c=2 you get one answer for x

OpenStudy (anonymous):

because it's a special case.

OpenStudy (anonymous):

okay I get all that now. But the reason way I asked this question is becaues I don't know how to set it up. like I dnt know which numbers to chose at the begining. That why I wanted some one to set up all three and now thanks to you I can do the rest.

OpenStudy (anonymous):

The thing is sometimes I have trouble too knowing how to set it up and Ive been taking math for a while now. Some people can see it really fast and some people can't. I can't so what I do is try to set up a complete square first. (x+2)(x+1) then I simplify it (x+2)(x+1)= x^2 + 3x + 2 then I see what a b and c equal

OpenStudy (anonymous):

That's how I get the general form of the equation. And if they tell me to complete the square I work backwards again.

OpenStudy (anonymous):

x^2 + 3x + 2=0 Complete the square Oh ok I can do it now It's (x+2)(x+1)

OpenStudy (anonymous):

(x+2)(x+1)=0 See how I did it.

OpenStudy (anonymous):

yah, I saw :)

OpenStudy (anonymous):

at least I know how to do that now.

OpenStudy (anonymous):

Cool I can do it faster now because I practice them a lot so if you keep doing it then sooner or later you will be able to catch them really fast.

OpenStudy (anonymous):

Good Luck.

OpenStudy (anonymous):

thanks , btw ur a gud teacher ;)

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