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Mathematics 14 Online
OpenStudy (anonymous):

Help 5 questions

OpenStudy (anonymous):

@nincompoop

OpenStudy (anonymous):

I could easily answer but I wouldn't know how to show my work xD wolframalpha.com

OpenStudy (anonymous):

I gave you a website to use~

OpenStudy (anonymous):

@zepdrix @Nurali @e.mccormick @agent0smith @amistre64 @nubeer @hba

OpenStudy (anonymous):

LxairaS.... I GAVE you a website that will give you the answer along with the step by step solution

OpenStudy (anonymous):

i couldnt find the correct answer

OpenStudy (anonymous):

for which question?

OpenStudy (anonymous):

nvm ill get help from @Nurali

OpenStudy (anonymous):

for #5: type in "factor 4x^2+5x-6" and it'll give you the direct answer...

OpenStudy (anonymous):

@e.mccormick can you help?

OpenStudy (e.mccormick):

Well, did you try factoring it?

OpenStudy (anonymous):

i think for 1 its 0 or a

OpenStudy (e.mccormick):

\(4x^2+5x-6 \) starts with a 4. The factors of 4 are 4 and 1 or 2 and 2. None of the answers starts with a 2, so you can skip looking at those.

OpenStudy (anonymous):

ohh okay 5 is b? i wa stalking about 1 lol

OpenStudy (e.mccormick):

It has a negative because the -6 at the end. The middle is +5, so the + part must be larger than the negative. There is also a formula for finding these but I don't remember it off the top of my head. However, with just those three pieces of information, I was able to find the factors.

OpenStudy (e.mccormick):

Ah, 1. Hehe. you were talking about 5 with the other guy. As for the first two, they are both discriminant problems. Did you use that part of the quadratic to find it?

OpenStudy (anonymous):

let me see if i can figure them out lol ill tell you when im done k :)

OpenStudy (e.mccormick):

\(b^2-4ac\) is all you need. From \(ax^2+bx+c\) form.

OpenStudy (anonymous):

1 is a 0 right ?

OpenStudy (anonymous):

@e.mccormick

OpenStudy (e.mccormick):

No.

OpenStudy (anonymous):

how?

OpenStudy (anonymous):

is it 32 right?

OpenStudy (anonymous):

i subtraced the 16s when i was supposed to add

OpenStudy (anonymous):

@e.mccormick i think 2 is either b or c

OpenStudy (e.mccormick):

\(4^2-4(2)(-2) = 16+16=23\) so yah.

OpenStudy (e.mccormick):

oops typoed there. 32.

OpenStudy (anonymous):

is 2 c 2 irrational solutions?

OpenStudy (e.mccormick):

\(4^2-4(2)(4) = 16-32 = -16\) Now, the \(\sqrt{-16}\) is that irrational, or is there another word for it?

OpenStudy (anonymous):

since -16 is not a perfect square?

OpenStudy (anonymous):

@e.mccormick ?

OpenStudy (e.mccormick):

Goes back to number systems. Look for \(\sqrt{-1}\) http://www.mathnstuff.com/math/spoken/here/2class/0/c0pg00.htm

OpenStudy (anonymous):

so its complex?

OpenStudy (e.mccormick):

Yah, whn a root is negative it is going to make what is called a pair of complex conjugates. Because of the \(\pm\) nature of roots they always come in pairs.

OpenStudy (anonymous):

okay for 3 my answer is c for 4 my answer is d for 5 my answer is b am i correct?

OpenStudy (e.mccormick):

5 is b. I have not done the others yet. Let me look....

OpenStudy (anonymous):

k

OpenStudy (e.mccormick):

3 yah.

OpenStudy (anonymous):

If I give e.mccormick a medal for showing work, will you give me one for answering the questions, lxairaS?

OpenStudy (e.mccormick):

4, yah d.

OpenStudy (anonymous):

yes give one to @e.mccormick

OpenStudy (anonymous):

@e.mccormick is 3 c @whalexnuker give a medal to @e.mccormick

OpenStudy (e.mccormick):

If there is ever a question of me vs someone else getting a medal, don't bother giving me one. I do not care about medals.

OpenStudy (anonymous):

ohh k is 3 c?

OpenStudy (e.mccormick):

Yes. That is what my 3 yah meant.

OpenStudy (anonymous):

Thanks @e.mccormick :)

OpenStudy (anonymous):

and @whalexnuker tahnks for the help :)

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