For the quadratic eqation `6x^2+11x+3=0`, if `alpha and beta` are the roots, then the value of `(6 alpha + 11)^2+(6 beta+11)^2` is

Answer

85

Explanation

since, `alpha and beta` are the roots of `6x^2+11x+3=0`
`implies 6alpha^2+11 alpha+3=0 and`
`6 beta^2+11 beta+3=0`
`implies 6 alpha +11 = (-3)/alpha and`
`6 beta+11=(-3)/beta`
`implies (6 alpha+11)^2+(6 beta+11)^2`
`=((-3)/alpha)^2+((-3)/beta)^2`
`9(1/alpha^2+1/beta^2)`
`9((alpha^2+beta^2)/(a^2beta^2))`
`=9[((alpha+beta^2)-2 alpha beta)/((alpha beta)^2)]`
`=9[((-11/6)^(2)-(1/2))/((1/2)^2)]`
`...[underset(alpha beta=3/6=1/2)("As" alpha+beta=(-11)/6)]`
`=9((121/36-1)/(1/4))`
`=85`