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Dec18-04, 07:06 AM   #1
Mo
 

Solve


I know how to solve quadratics using both factorisation and the equation method ... but how can i solve :

[tex](x-1)(2x-1)(3x-1) = 0 [/tex]

I multiplied it all out and i got ..

[tex]6x^3 - 2x^2 -3x -1=0[/tex]

I just do not know where to got from here .. a little nudge in the right direction would be appreciated!

regards,
Mo
 
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Dec18-04, 07:13 AM   #2
 
Quote by Mo
I know how to solve quadratics using both factorisation and the equation method ... but how can i solve :

[tex](x-1)(2x-1)(3x-1) = 0 [/tex]

I multiplied it all out and i got ..

[tex]6x^3 - 2x^2 -3x -1=0[/tex]

I just do not know where to got from here .. a little nudge in the right direction would be appreciated!

regards,
Mo
A small hint: why do you multiply? After all, if a*b*c=0, then....
 
Dec18-04, 07:13 AM   #3
 
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Quote by Mo
I know how to solve quadratics using both factorisation and the equation method ... but how can i solve :
[tex](x-1)(2x-1)(3x-1) = 0 [/tex]
I multiplied it all out and i got ..
[tex]6x^3 - 2x^2 -3x -1=0[/tex]
I just do not know where to got from here .. a little nudge in the right direction would be appreciated!
regards,
Mo
Why the heck did u multiply it??? It was already solved.U were being asked for the 3 possible values of "x" which cancel the expression from the LHS.I think/hope they were obvious...

Daniel.

PS.But if u'd rather apply Cardano's formulae for the 3rd order algebrac equation u got,be my guest...
 
Dec18-04, 07:20 AM   #4
Mo
 

Solve




yes, i do understand.

if

(x-1) = 0 then x =1
(2x-1) =0 then 2x=1 so x = 1/2
(3x-1) = 0 then 3x=1 so x=1/3

is that right.Am i going about it right. thanks.
 
Dec18-04, 07:27 AM   #5
 
Yeah, that's right.
 
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