If `alpha!=betaa n dalpha^2=5alpha-3a n dbeta^2=5beta-3.` find the equation whose roots are `alpha//betaa n dbeta//alphadot`

Explanation

We have `alpha(2) = 5alpha - 3 and beta^(2) = 5beta - 3.` Hence, `alpha, beta` are roots of `x^(2) = 5x - 3, ie., x^(2) - 5x + 3 = 0` . Therefore
`alpha + beta = 5 and alpha beta = 3`
Now, `S=(alpha)/(beta)+(beta)/(alpha) = (alpha^(2) + beta^(2))/(alphabeta)`
`=((alpha+beta)^(2)-2alphabeta)/(alpha beta) = (25 - 6)/(3) = (19)/(3)`
and `P=(alpha)/(beta)(beta)/(alpha)=1`
So the required equation is
`x^(2) - Sx + P = 0`
`rArr x^(2) - (19)/(3) x + 1 = 0`
or ` 3x^(2) - 19x + 3 = 0`