Question

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

Answer 1

\[\huge 0 = -4(-4t^2 -8t -1)\]

Answer 2

Well the first step is to compare your equation with the standard form and get a,b and c

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

Answer 3

So, how would I do that with this one? Multiply the parentheses stuff by -4?

Answer 4

Seeing your question i would suggest you to first divide your whole equation by -4 and then compare

Answer 5

\[\Huge 0=-4t^2 -8t-1\]

Answer 6

What is a,b and c in this ?

Answer 7

a=-4 b=-8 c=-1

Answer 8

Good,Now you can use the quad formula.

\[\huge\ x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 9

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 :)

Answer 10

so, t = -8*√64-4*-4
------------------
2*-4 ?

Answer 11

which would then be t=-8*(8+4)
-----------
-8

Answer 12

am I getting it right?

Answer 13

It should be +- not *

Answer 14

I don't get it... wouldn't you be multiplying -8 and \[\pm √64-4*-4\]?

Answer 15

t=-8(+-)(8+4)
-----------
-8

Answer 16

Now I'm rather confused.

Answer 17

Lemme show you.

Answer 18

I appreciate you working with me though.

Answer 19

I guess I could do that \(\Huge\mathsf{:P}\)

Answer 20

a=-4 b=-8 c=-1

\[\huge\ t=\frac{ -(-8) \pm \sqrt{(-8)^2-4(-4)(-1)} }{ 2(-4) }\]

Answer 21

Ok, but what does the \[\pm\] do? what's its significance?

Answer 22

Well when we simplify it we would be taking the adding part on one side and subracting part on the other side.

Answer 23

so it'd be 8+√blahblahblah
---------------
2-(-4)?

Answer 24

Either,

8+√blahblahblah
---------------
2-(-4)?

Or,
8-√blahblahblah
---------------
2-(-4)?

Answer 25

so the \[\pm\] just breaks up between the denominator and the numerator?

Answer 26

Exactly.

Answer 27

Ah. I guess I'll try another problem here and ask you if it's right or something.

Answer 28

Sure :)

Answer 29

\[\huge 0=-2x ^{2} +14x +20\]
\[\large x=\frac{ -14\pm \sqrt{14^2 - 4(-2+20)} }{ -4 }\]
\[\large x= \frac{ -14+14-72 }{ -4 }\]

Answer 30

and the answer is

\[\huge \frac{ -7\pm \sqrt{89} }{ 2 }\]. Would I be right?

Answer 31

Its is -4(a)(c) not -4(a+c)

Answer 32

oh. Well that ruins that then.

Answer 33

Yes it does :)


Do it on a paper and tell me what you got from that equation :)

Answer 34

\[\Large x=\frac{ -14\pm \sqrt{14^2 - 4(-2*20)} }{ -4 }\]
\[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\]
\[\Large x=\frac{ -14\pm 14-160} { -4}\]
\[\Large x=\frac{ -14\pm -146} {-4}\]
set x to 0 and divide all by -2 to get \[\Large \frac{ 7\pm 73} { 2}\]

Answer 35

and now I'm confused.

Answer 36

oh wait, I forgot to keep the x's in there. Crap.

Answer 37

Think you could help me out here? My mind is about to implode.

Answer 38

x=7+73/2,x=7-73/2

Answer 39

Btw,

(-) *(-)=+

Answer 40

So just because of signs it goes wrong.

Answer 41

\[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\]

Continue from here.

Answer 42

But I left the xs out of my work. ._.

Answer 43

No worries about that :)

Answer 44

Continue working from here.

Answer 45

\[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\]
\[\Large x=\frac{ -14\pm 14+160} {-4}\]
\[\Large x=\frac{ -14\pm 174} {-4}\]
\[\Large x=\frac {\frac{ -14\pm 174} {-4}} {-2}\]
\[\Large x=\frac {7\pm -87} {2}\]

Answer 46

And now what do I do?

Answer 47

I don't know what you are upto ,where did that divide by -2 come in between ?

Answer 48

Well, I'm not sure. I just thought I had to do that. I'm rather confused ._.

Answer 49

Hmm,If OS would be working on chrome i would have worked one problem for you.

Can we do this later ?

Answer 50

I'm using chrome right now o_o But, I guess that'd be OK :P

Answer 51

Thanks.

Answer 52

Thanks to you too :3