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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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???
looks fun but incomplete.
its complete

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add my fb if u wanna get down ;) https://www.facebook.com/ParthKohli
@mathmath333 is this equation really having 3 real roots??
\(\large \color{black}{\begin{align} \text{if}\ p,q,r\ \text{are roots of }\hspace{.33em}\\~\\ 2x^3-3x^2-x-1=0 \hspace{.33em}\\~\\ \text{Find}\ (1-p)(1-q)(1-r) \end{align}}\)
look at the degree of x , it will have 3 roots
are u talking about 3 real roots???
no
ok :)
is the answer -1/2
no
-3/2
how ?
yes -3/2 is correct
(x-p)(x-q)(x-r) = 0 multiply that out gives x^2 - 2x^2 (p + q + r) + 2x (qr + pq + pr) - 2pqr = 0 pattern match to original equation so times 2 multiply out (1-p)(1-q)(1-r) and you get same patterns eg p + q + r, (qr + pq + pr) , pqr and pattern match long winded
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\(\large \color{black}{\begin{align} \ &2x^3-3x^2-x-1=0 \hspace{.33em}\\~\\ &\implies x^3-\dfrac{3}{2}x^2-\dfrac{x}{2}-\dfrac{1}{2}=0 \hspace{.33em}\\~\\ &(x-p)(x-q)(x-q)=x^3-\dfrac{3}{2}x^2-\dfrac{x}{2}-\dfrac{1}{2}\hspace{.33em}\\~\\ &\text{put} \quad x=1 \hspace{.33em}\\~\\ &(1-p)(1-q)(1-q)=1^3-\dfrac{3}{2}\times 1^2-\dfrac{1}{2}-\dfrac{1}{2}\hspace{.33em}\\~\\ \end{align}}\)
aaargh! much better
i will put up another such question
:)
This is simply transformation of roots, isn't it?
If \(a_1, a_2 , \cdots , a_n\) are the roots of \(p_n(x)=0\) where \(n\) is the degree of the polynomial, then \(1 - a_1, 1 - a_2, \cdots, 1 - a_n\) are the roots of \(p_n(1 - x)=0\).
\[p'(x) = p(1-x) = 2(1-x)^3 - 3(1 - x)^2 - (1 - x) - 1\]
\[= -2 x^3+3 x^2+x-3\]The product of roots of this \(-3/2\).

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