Question

Fun question

Answer 1

???

Answer 2

looks fun but incomplete.

Answer 3

its complete

Answer 4

add my fb if u wanna get down ;) https://www.facebook.com/ParthKohli

Answer 5

@mathmath333 is this equation really having 3 real roots??

Answer 6

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

Answer 7

look at the degree of x , it will have 3 roots

Answer 8

are u talking about 3 real roots???

Answer 9

no

Answer 10

ok :)

Answer 11

is the answer -1/2

Answer 12

no

Answer 13

-3/2

Answer 14

how ?

Answer 15

yes -3/2 is correct

Answer 16

(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

Answer 17

\(\large \color{black}{\begin{align} \
&2x^3-3x^2-x-1=0 \hspace{.33em}\\~\\
&\implies x^3-\dfrac{3}{2}x^2-\dfrac{x}{2}-\dfrac{1}{2}=0 \hspace{.33em}\\~\\
&(x-p)(x-q)(x-q)=x^3-\dfrac{3}{2}x^2-\dfrac{x}{2}-\dfrac{1}{2}\hspace{.33em}\\~\\
&\text{put} \quad x=1 \hspace{.33em}\\~\\
&(1-p)(1-q)(1-q)=1^3-\dfrac{3}{2}\times 1^2-\dfrac{1}{2}-\dfrac{1}{2}\hspace{.33em}\\~\\
\end{align}}\)

Answer 18

aaargh! much better

Answer 19

i will put up another such question

Answer 20

:)

Answer 21

This is simply transformation of roots, isn't it?

Answer 22

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 \(p_n(1 - x)=0\).

Answer 23

\[p'(x) = p(1-x) = 2(1-x)^3 - 3(1 - x)^2 - (1 - x) - 1\]

Answer 24

\[= -2 x^3+3 x^2+x-3\]The product of roots of this \(-3/2\).