Question

Can someone please help with the attached question?

Answer 1

\[ x^3 +4x - 1 = \left( x - \alpha\right)\left( x - \beta\right)\left( x - \gamma\right) = \left( \frac{1 - y}{y} - \alpha\right)\left( \frac{1 - y}{y} - \beta\right)\left( \frac{1 - y}{y} - \gamma\right ) = 0\]

Show that
\[ \frac{1 - y}{y} - \alpha = 0 \implies y = \frac1{1 + \alpha } \]
do rest for others.

Answer 2

the sum of roots of polynomial of degree n should be equal to negative coefficient of x^(n-1)
\[ \frac{1}{1 + \alpha} + \frac{1}{1 + \beta} + \frac{1}{1 + \gamma} = 7\]

Answer 3

\[ \left( \frac{1}{1 + \alpha} + \frac{1}{1 + \beta} + \frac{1}{1 + \gamma} \right)^2 = 49 \]
\[ (a + b + c)^2 = a^2 + b^2 + c^2 - 2 (ab +bc + ac)\]
and \( (ab +bc + ac) = 3\) .. use this technique to evaluate the case when n=2

Answer 4

Woops!! ... forgot that there is 6 in front of y^3 ... you should be dividing it by 6 before comparing.

Answer 5

Thank you so much! :)
Could you please show me how to do the last part (Hence show that....)?

Answer 6

\[ 6y^3-7y^2+3y-1=0\implies 6y^3= -(-7y^2+3y-1) \\
\text{ put }y ={1 \over 1+\alpha}\]
since you know the values of ^2's and ^1's ... you can get it too.

Answer 7

for the last part
\[ \sum {(\beta + 1)(\gamma +1) \over (\alpha + 1)^2 }=\sum {(\alpha +1)(\beta + 1)(\gamma +1) \over (\alpha + 1)^3 }\]
\[ (\alpha +1)(\beta + 1)(\gamma +1) = \text{ product of roots} = {-1 \over 6}\]
put the value of cubes calculated from above post.

Answer 8

I wish you were my teacher! ^_^

Answer 9

lol ... not really.

Answer 10

Haha. Why? Would you be a strict one? :P

Answer 11

no ... i'm not strict and students would never study and fail.