`2x^(2) + 3x - alpha - 0 " has roots "-2 and beta " while the equation "x^(2) - 3mx + 2m^(2) = 0 " has both roots positive, where " alpha gt 0 and beta gt 0.`
If `alpha and beta ` are the roots of the equation `3x^(2) + 2 x + 1 = 0 ,` then the equation whose roots are `alpha + beta ^(-1) and beta + alpha ^(-1)`

Options

A. `3x^(2) + 8x + 16 = 0 `
B. `3x^(2) - 8x - 16 = 0`
C. `3x^(2) + 8x - 16 = 0 `
D. `x^(2) + 8x + 16 = 0`

Answer

A

Explanation

`3x^(2) + 2x+1=0`
Sum of the roots ` alpha + beta = (-2)/3" "`…(1)
Product of the roots ` = alpha *beta = 1/3" "` ….(2)
We have to find the equation with roots `alpha + beta^(-1) and beta+alpha^(-1)`
Sum of the roots (S) `= alpha + beta^(-1) + beta+ alpha^(-1)`
`=alpha +1/beta + beta + 1/alpha`
` = (alpha + beta) + ((alpha+ beta)/(alphabeta))`
` = (-2)/3+((-2/3)/(1/3))" "` (from (1), (2))
` =-2/3 - 2 = (-2-6)/3 = -8/3`.
Product of the roots (P) `= (alpha + beta^(-1))(beta + alpha^(-1))`
` = (alpha + 1/beta)(beta+1/alpha)`
` alpha beta + 1 + 1+ 1/(alpha beta) = alpha beta + 2 + 1/(alpha beta)`
` 1/3 + 2 + 1/(1/3)` ,
`= 1/3 + 2 + 3 =1/3 + 5 = 16/3 . `
So, the required equation is `x^(2) - s.x. + P = 0 `
` = x^(2) + 8/3x + 16/3 = 0 rArr 3x^(2) + 8x + 16 = 0 `