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Can someone help explain how to do quadratic equations with the quadratic formula? I'll give you a problem. See comments.

Mathematics
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\[\huge 0 = -4(-4t^2 -8t -1)\]
  • hba
Well the first step is to compare your equation with the standard form and get a,b and c \[\huge\ ax^2+bx+c=0\]
So, how would I do that with this one? Multiply the parentheses stuff by -4?

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Other answers:

  • hba
Seeing your question i would suggest you to first divide your whole equation by -4 and then compare
\[\Huge 0=-4t^2 -8t-1\]
  • hba
What is a,b and c in this ?
a=-4 b=-8 c=-1
  • hba
Good,Now you can use the quad formula. \[\huge\ x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
  • hba
The other think which you can do with your equation is that you can take -1 common and then compare so everything becomes positive.you can also do it straight away :)
so, t = -8*√64-4*-4 ------------------ 2*-4 ?
which would then be t=-8*(8+4) ----------- -8
am I getting it right?
  • hba
It should be +- not *
I don't get it... wouldn't you be multiplying -8 and \[\pm √64-4*-4\]?
  • hba
t=-8(+-)(8+4) ----------- -8
Now I'm rather confused.
  • hba
Lemme show you.
  • hba
I appreciate you working with me though.
I guess I could do that \(\Huge\mathsf{:P}\)
  • hba
a=-4 b=-8 c=-1 \[\huge\ t=\frac{ -(-8) \pm \sqrt{(-8)^2-4(-4)(-1)} }{ 2(-4) }\]
Ok, but what does the \[\pm\] do? what's its significance?
  • hba
Well when we simplify it we would be taking the adding part on one side and subracting part on the other side.
so it'd be 8+√blahblahblah --------------- 2-(-4)?
  • hba
Either, 8+√blahblahblah --------------- 2-(-4)? Or, 8-√blahblahblah --------------- 2-(-4)?
so the \[\pm\] just breaks up between the denominator and the numerator?
  • hba
Exactly.
Ah. I guess I'll try another problem here and ask you if it's right or something.
  • hba
Sure :)
\[\huge 0=-2x ^{2} +14x +20\] \[\large x=\frac{ -14\pm \sqrt{14^2 - 4(-2+20)} }{ -4 }\] \[\large x= \frac{ -14+14-72 }{ -4 }\]
and the answer is \[\huge \frac{ -7\pm \sqrt{89} }{ 2 }\]. Would I be right?
  • hba
Its is -4(a)(c) not -4(a+c)
oh. Well that ruins that then.
  • hba
Yes it does :) Do it on a paper and tell me what you got from that equation :)
\[\Large x=\frac{ -14\pm \sqrt{14^2 - 4(-2*20)} }{ -4 }\] \[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\] \[\Large x=\frac{ -14\pm 14-160} { -4}\] \[\Large x=\frac{ -14\pm -146} {-4}\] set x to 0 and divide all by -2 to get \[\Large \frac{ 7\pm 73} { 2}\]
and now I'm confused.
oh wait, I forgot to keep the x's in there. Crap.
Think you could help me out here? My mind is about to implode.
  • hba
x=7+73/2,x=7-73/2
  • hba
Btw, (-) *(-)=+
  • hba
So just because of signs it goes wrong.
  • hba
\[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\] Continue from here.
But I left the xs out of my work. ._.
  • hba
No worries about that :)
  • hba
Continue working from here.
\[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\] \[\Large x=\frac{ -14\pm 14+160} {-4}\] \[\Large x=\frac{ -14\pm 174} {-4}\] \[\Large x=\frac {\frac{ -14\pm 174} {-4}} {-2}\] \[\Large x=\frac {7\pm -87} {2}\]
And now what do I do?
  • hba
I don't know what you are upto ,where did that divide by -2 come in between ?
Well, I'm not sure. I just thought I had to do that. I'm rather confused ._.
  • hba
Hmm,If OS would be working on chrome i would have worked one problem for you. Can we do this later ?
I'm using chrome right now o_o But, I guess that'd be OK :P
  • hba
Thanks.
Thanks to you too :3

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