how does this: (3 - &)(2-&)(4-&) - 2(8-2&)-2(4-2&)= 0 where & is equal to lambada(sorry cant seem to get symbol on) end up as this: &^3 - 9&^2 + 18& = 0 cheers
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Originally Posted by thermalwarrior how does this: (3 - &)(2-&)(4-&) - 2(8-2&)-2(4-2&)= 0 where & is equal to lambada(sorry cant seem to get symbol on) end up as this: &^3 - 9&^2 + 18& = 0 cheers Small wonder that you are having troubles: It doesn't. Your expression comes out as $\displaystyle -\lambda ^3 + \lambda ^2 + 22 \lambda - 40$ -Dan
Hello, Originally Posted by topsquark Small wonder that you are having troubles: It doesn't. Your expression comes out as $\displaystyle -\lambda ^3 + \lambda ^2 + 22 \lambda - 40$ -Dan I find it to be $\displaystyle -\lambda^3+9 \lambda^2-18 \lambda=0$, simply by developping This equation is the one given (divide by -1).
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