Let `alpha`, `beta` be two real numbers satisfying the following relations `alpha^(2)+beta^(2)=5`, `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))`
Possible value of `alpha + beta` is

Options

A. `x^(2)+-x+2=0`
B. `x^(2)+-3x-2=0`
C. `x^(2)+-sqrt(3)x+2=0`
D. none of these

Answer

D

Explanation

`alpha^(2)+beta^(2)=5`
`3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))`
`(alpha^(5)+beta^(5))/(alpha^(3)+beta^(3))=(11)/(3)`
` :. ((alpha^(3)+beta^(3))(alpha^(2)+beta^(2))-(alpha^(2)beta^(2)(alpha+beta)))/(alpha^(3)+beta^(3))=(11)/(3)`
`:. alpha^(2)+beta^(2)-(alpha^(2)beta^(2)(alpha+beta))/((alpha+beta)(alpha^(2)+beta^(2)-alphabeta))=(11)/(3)`
` :. 5-(alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` ltbrlt `:. (25-5alphabeta-alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)`
Let `alphabeta=t`
`(25-5t-t^(2))/(5-t)=(11)/(3)`
`75-15t-3t^(2)=55-11 t`
`75-15t-3t^(2)-55+11t=0`
`-3t^(2)-4t+20=0`
`(t-2)(3t+10)=0`
` :. t=2` or `(-10)/(3)`
So `alpha beta=2`, `alphabeta=(-10)/(3)`
If `alphabeta=2`,
`alpha^(2)+beta^(2)=(alpha+beta)^(2)-2alphabeta`
`:.5=(alpha+beta)^(2)-2xx2`
`(alpha+beta)^(2)=9`
`alpha+beta=+-3`
for `alphabeta=(-10)/(3)`, `(alpha+beta)^(2) lt 0`
`implies x^(2) +- 3x+2=0`