Question

If alpha and beta are the zeroes of 6x2 + x-1 find the values of alpha3beta + alphabeta

Answer 1

6x²+x-1 = 0

⇒6x²+3x-2x-1=0

⇒(3x-1)(2x+1)=0

⇒(α=1/3 and β=-1/2) or (α=-1/2 and β=1/3)

So α³β+αβ

⇒(1/3)³(-1/2)+(1/3)(-1/2)   or   (-1/2)³(1/3)+(-1/2)(1/3)

⇒-5/27 or -5/24