mathmath333
  • mathmath333
Fun question
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jhonyy9
  • jhonyy9
???
anonymous
  • anonymous
looks fun but incomplete.
mathmath333
  • mathmath333
its complete

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
add my fb if u wanna get down ;) https://www.facebook.com/ParthKohli
imqwerty
  • imqwerty
@mathmath333 is this equation really having 3 real roots??
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \text{if}\ p,q,r\ \text{are roots of }\hspace{.33em}\\~\\ 2x^3-3x^2-x-1=0 \hspace{.33em}\\~\\ \text{Find}\ (1-p)(1-q)(1-r) \end{align}}\)
mathmath333
  • mathmath333
look at the degree of x , it will have 3 roots
imqwerty
  • imqwerty
are u talking about 3 real roots???
mathmath333
  • mathmath333
no
imqwerty
  • imqwerty
ok :)
imqwerty
  • imqwerty
is the answer -1/2
mathmath333
  • mathmath333
no
IrishBoy123
  • IrishBoy123
-3/2
mathmath333
  • mathmath333
how ?
imqwerty
  • imqwerty
yes -3/2 is correct
IrishBoy123
  • IrishBoy123
(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
IrishBoy123
  • IrishBoy123
1 Attachment
imqwerty
  • imqwerty
1 Attachment
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \ &2x^3-3x^2-x-1=0 \hspace{.33em}\\~\\ &\implies x^3-\dfrac{3}{2}x^2-\dfrac{x}{2}-\dfrac{1}{2}=0 \hspace{.33em}\\~\\ &(x-p)(x-q)(x-q)=x^3-\dfrac{3}{2}x^2-\dfrac{x}{2}-\dfrac{1}{2}\hspace{.33em}\\~\\ &\text{put} \quad x=1 \hspace{.33em}\\~\\ &(1-p)(1-q)(1-q)=1^3-\dfrac{3}{2}\times 1^2-\dfrac{1}{2}-\dfrac{1}{2}\hspace{.33em}\\~\\ \end{align}}\)
IrishBoy123
  • IrishBoy123
aaargh! much better
mathmath333
  • mathmath333
i will put up another such question
imqwerty
  • imqwerty
:)
ParthKohli
  • ParthKohli
This is simply transformation of roots, isn't it?
ParthKohli
  • ParthKohli
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 \(p_n(1 - x)=0\).
ParthKohli
  • ParthKohli
\[p'(x) = p(1-x) = 2(1-x)^3 - 3(1 - x)^2 - (1 - x) - 1\]
ParthKohli
  • ParthKohli
\[= -2 x^3+3 x^2+x-3\]The product of roots of this \(-3/2\).

Looking for something else?

Not the answer you are looking for? Search for more explanations.