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Mathematics 22 Online
OpenStudy (anonymous):

Help for a medal???

OpenStudy (anonymous):

If -nx2 + tx + c = 0, what is x equal to?

OpenStudy (solomonzelman):

you have to complete the square for the left side. (or I can refer to that "find the vertex form of the left side" )

OpenStudy (anonymous):

But how would I do that?

OpenStudy (solomonzelman):

Ok, at first you need to factor the first two terms out of -n

OpenStudy (anonymous):

But you see. The thing is, I don't know how to factor when its mostly letters

OpenStudy (anonymous):

It really confuses me when its letters and not numbers

OpenStudy (solomonzelman):

if you know how to factor when you have numbers, it is a good place to start at then

OpenStudy (solomonzelman):

\(\large\color{black}{ \displaystyle -nx^2 + tx + c = 0 }\) would you agree with what I did below? \(\large\color{black}{ \displaystyle -nx^2 + \frac{-ntx}{-n} + c = 0 }\)

OpenStudy (anonymous):

Yup

OpenStudy (solomonzelman):

ok, now, each term has -n in it (each of the first 2). take that -n out from the first term.

OpenStudy (misty1212):

i have an idea use \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] only where you see an \(a\) put\(-n\) , where you see a \(b\) put a \(t\) and leave \(c\) as it is

OpenStudy (solomonzelman):

lol. I always there with completing the square. Quad form is definitely better. Nice approach!

OpenStudy (anonymous):

Alright, thanks you two! By the way misty, LOVE THE HAIR!

OpenStudy (solomonzelman):

Do you like my hair ? :)

OpenStudy (solomonzelman):

OH, don't tell me I am bold, you will offend me.

OpenStudy (anonymous):

Haha, yeah! Very purple-ish

OpenStudy (misty1212):

\[\huge \color\magenta\heartsuit\]

OpenStudy (solomonzelman):

have a nice evening ladies. Adios

OpenStudy (anonymous):

If I knew how to make that heart I would type it right now. <3 That should be good enough though

OpenStudy (anonymous):

Thanks, you too!

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