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mathmath333

  • one year ago

Fun question

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  1. jhonyy9
    • one year ago
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    ???

  2. anonymous
    • one year ago
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    looks fun but incomplete.

  3. mathmath333
    • one year ago
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    its complete

  4. anonymous
    • one year ago
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    add my fb if u wanna get down ;) https://www.facebook.com/ParthKohli

  5. imqwerty
    • one year ago
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    @mathmath333 is this equation really having 3 real roots??

  6. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \text{if}\ p,q,r\ \text{are roots of }\hspace{.33em}\\~\\ 2x^3-3x^2-x-1=0 \hspace{.33em}\\~\\ \text{Find}\ (1-p)(1-q)(1-r) \end{align}}\)

  7. mathmath333
    • one year ago
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    look at the degree of x , it will have 3 roots

  8. imqwerty
    • one year ago
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    are u talking about 3 real roots???

  9. mathmath333
    • one year ago
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    no

  10. imqwerty
    • one year ago
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    ok :)

  11. imqwerty
    • one year ago
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    is the answer -1/2

  12. mathmath333
    • one year ago
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    no

  13. IrishBoy123
    • one year ago
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    -3/2

  14. mathmath333
    • one year ago
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    how ?

  15. imqwerty
    • one year ago
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    yes -3/2 is correct

  16. IrishBoy123
    • one year ago
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    (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

  17. IrishBoy123
    • one year ago
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  18. imqwerty
    • one year ago
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  19. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \ &2x^3-3x^2-x-1=0 \hspace{.33em}\\~\\ &\implies x^3-\dfrac{3}{2}x^2-\dfrac{x}{2}-\dfrac{1}{2}=0 \hspace{.33em}\\~\\ &(x-p)(x-q)(x-q)=x^3-\dfrac{3}{2}x^2-\dfrac{x}{2}-\dfrac{1}{2}\hspace{.33em}\\~\\ &\text{put} \quad x=1 \hspace{.33em}\\~\\ &(1-p)(1-q)(1-q)=1^3-\dfrac{3}{2}\times 1^2-\dfrac{1}{2}-\dfrac{1}{2}\hspace{.33em}\\~\\ \end{align}}\)

  20. IrishBoy123
    • one year ago
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    aaargh! much better

  21. mathmath333
    • one year ago
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    i will put up another such question

  22. imqwerty
    • one year ago
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    :)

  23. ParthKohli
    • one year ago
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    This is simply transformation of roots, isn't it?

  24. ParthKohli
    • one year ago
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    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 \(p_n(1 - x)=0\).

  25. ParthKohli
    • one year ago
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    \[p'(x) = p(1-x) = 2(1-x)^3 - 3(1 - x)^2 - (1 - x) - 1\]

  26. ParthKohli
    • one year ago
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    \[= -2 x^3+3 x^2+x-3\]The product of roots of this \(-3/2\).

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