Question

If `alpha,beta` are the roots of the equation `2x^2-35+2=0` , the find the value of `(2alpha-35)^3(2beta-35)^3dot`

Answer 1

Since `alpha, beta ` are the roots of the equation, we have
`2x^(2) - 35x + 2 = 0`
`therefore 2a^(2) - 35alpha = - 2 or 2 alpha - 35 = (-2)/(alpha)`
and `2beta^(2) - 35 beta = - 2 or 2 beta -35 = (-2)/(beta)`
Now `(2alpha - 35 )^(3) (2beta - 35)^(3) = ((-2)/(alpha))((-2)/(beta))^(3)`
`(8xx8)/(alpha^(3)beta^(3) )=(64)/(1) = 64" " (thereforealphabeta = 1)`