If [ ] and [ ] are the roots of the equation x2 + px + 2 = 0 and [1 ] and [1 ] are the roo…

Question

If $$\alpha $$ and $$\beta $$ are the roots of the equation

x2
 + px + 2 = 0 and $${1 \over \alpha }$$ and $${1 \over \beta }$$ are the
 roots of
the equation 2x2
 + 2qx + 1 = 0, then

$$\left( {\alpha  - {1 \over \alpha }} \right)\left( {\beta  - {1 \over \beta }} \right)\left( {\alpha  + {1 \over \beta }} \right)\left( {\beta  + {1 \over \alpha }} \right)$$ is equal to :

Accepted Answer

$$\alpha $$ and $$\beta $$ are the roots of the

equation
x2
 + px + 2 = 0

$$ \therefore $$ $$\alpha  + \beta  =  - p,\,\alpha \beta  = 2$$

$${1 \over \alpha }$$ and $${1 \over \beta }$$ are the roots of
the

equation 2x2
 + 2qx + 1 = 0

$$ \therefore $$ $${1 \over \alpha } + {1 \over \beta } =  - q,\,{1 \over {\alpha \beta }} = {1 \over 2}$$

$$ \Rightarrow $$ $${{\alpha  + \beta } \over {\alpha \beta }} =  - q \Rightarrow {{ - p} \over 2} =  - q$$

$$ \Rightarrow p = 2q$$

$$\left( {\alpha  + {1 \over \beta }} \right)\left( {\beta  + {1 \over \alpha }} \right) = \alpha \beta  + {1 \over {\alpha \beta }} + 2$$ $$ = 2 + {1 \over 2} + 2 = {9 \over 2}$$

$$\left( {\alpha  - {1 \over \alpha }} \right)\left( {\beta  - {1 \over \beta }} \right) = \alpha \beta  + {1 \over {\alpha \beta }} - {\alpha  \over \beta } - {\beta  \over \alpha }$$

$$ = 2 + {1 \over 2} - \left[ {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right]$$ $

= {5 \over 2} - \left[ {{{{{\left( {\alpha  + \beta } \right)}^2} - 2\alpha \beta } \over {\alpha \beta }}} \right]$$ = {5 \over 2} - \left[ {{{{p^2} - 4} \over 2}} \right]$$ = {{9 - {p^2}} \over 2}$$\left( {\alpha  - {1 \over \alpha }} \right)\left( {\beta  - {1 \over \beta }} \right)\left( {\alpha  + {1 \over \beta }} \right)\left( {\beta  + {1 \over \alpha }} \right) = \left( {{{9 - {p^2}} \over 2}} \right)\left( {{9 \over 2}} \right) (

\] = {9 \over 4}(9 - {p^2})$

Solutions · 1