Since `alpha, beta ` are roots of the equation `2x^(2) – 3x – 6 = 0, ` we have
`alpha + beta = 3//2 and alpha beta = -3`
`rArr alpha^(2) + beta^(2) = (alpha + beta)^(2) – 2alpha beta`
= `(90/(4)+6 = (33)/(4)`
Now , `(alpha^(2)+beta^(2))+(beta^(2)+2) = (alpha^(2)+ beta^(2)) + 4`
`=(33)/(4) + 4 = (49)/(4)`
and `(alpha ^(2) + 2 ) (beta^(2)+2)= alpha^(2) + beta^(2) + 2(alpha^(2)+ beta^(2))+4`
`=(-3)^(2) + 2((33)/(4)) + 4`
`= (59)/(2)`
So , the equatioon whose roots are `alpha^(2) + 2 and beta^(2) + 2` is given by
`x^(2) – x [(alpha^(2) + 2) + (beta^(2) + 2)] + (alpha^(2) + 2 ) (beta^(2) + 2 ) = 0`
`rArr x^(2) – (49)/(4)x+(59)/(2) = 0`
or `4x^(2) – 49x + 118 = 0`