Math problem! - BasilMarket Forum

Math problem!

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thread starter eKyl Aug 30 2009 + PM | QUOTE | PERMALINK | REPORT

So I've just started Math C in Gymnasium... and I've encountered a problem on the equation-chapter. The equation is this: 4(3-3x)(8-2x�) = 0 How do I solve it? I've already made some progress on it... but I'd like to see how you solve it, because what I've done so far... might be wrong. Edit: Oh btw, if you know how to solve it, please explain in details!

Boondock5 08/30/09 + PM | QUOTE | PERMALINK | REPORT

FOIL it? I dunno. I forget math during summer.

USFRanger 08/30/09 + PM | QUOTE | PERMALINK | REPORT

Remember, brackets first. First, simplify the brackets, then move the sold numbers to the other side of the equal sign.

KimKibum 08/30/09 + PM | QUOTE | PERMALINK | REPORT

i think i knw how to do it

peterisnoob Aug 30 2009 + PM | QUOTE | PERMALINK | REPORT

woops meant 1, typed 0 xD x^3-x^2-2x+2=0 x^3-x^2-2x=-2 x(x^2-x-2)=-2

Floob 08/30/09 + PM | QUOTE | PERMALINK | REPORT

x = -2, x = 1, x = 2.

FENRIR82 08/30/09 + PM | QUOTE | PERMALINK | REPORT

Dont solve brackets! You have to find what values for X would make any term 0 (that what the result can be a total 0)... so X can be 1 (making firt term 0) or 2 (making second term 0)...every cuadratic equation have more than one answer (in most cases)

jiggoha 08/30/09 + PM | QUOTE | PERMALINK | REPORT

Floob : x = -2, x = 1, x = 2. That's what I got.

Icemaggi 08/30/09 + PM | QUOTE | PERMALINK | REPORT

http://www.dvbs-online.de/img_liese/4_1.jpg this?

FENRIR82 08/30/09 + PM | QUOTE | PERMALINK | REPORT

Floob is right... -2 is also a possible answer... my bad

eKyl 08/30/09 + PM | QUOTE | PERMALINK | REPORT

Floob : x = -2, x = 1, x = 2. Yeah, but how? :O Explain in details! Step by step :P

Unvalid 08/30/09 + PM | QUOTE | PERMALINK | REPORT

I believe this is correct (photobucket) Since Floob and Jiggoha got the same as I did

kenshin2020 08/30/09 + PM | QUOTE | PERMALINK | REPORT

Unvalid : I believe this is correct (photobucket) Since Floob and Jiggoha got the same as I did Geez, why take the hard way o.o See, since its 4(3-3x)(8-2x^2)=0, the key here is ALL = 0, so one of the bracket must have a total of 0 to get a multiplication result of 0. Either (3-3x) = 0 or (8-2x^2) =0. Solve each of those 2 equation separately, and u'll get one answer from the linear and two answers from the quadratic.

dattyboy4916 08/30/09 + PM | QUOTE | PERMALINK | REPORT

kenshin2020 : Geez, why take the hard way o.o See, since its 4(3-3x)(8-2x^2)=0, the key here is ALL = 0, so one of the bracket must have a total of 0 to get a multiplication result of 0. Either (3-3x) = 0 or (8-2x^2) =0. Solve each of those 2 equation separately, and u'll get one answer from the linear and two answers from the quadratic. Deeeep. I would've mindlessly foiled.

eKyl 08/30/09 + PM | QUOTE | PERMALINK | REPORT

Okay thanks! ( Special thanks to Unvalid ) I understand much better now ^^