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Solve |
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| Dec18-04, 07:06 AM | #1 |
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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 |
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| Dec18-04, 07:13 AM | #3 |
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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...
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| Dec18-04, 07:20 AM | #4 |
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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 |
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Yeah, that's right.
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