Question

Ans: 8 Let ( alpha ) and ( beta ) are zeroes of quadratic polynomial ( p(s)=3 s^{2} ) ( -6 s+4 )
So, ( alpha+beta=-left(frac{-6}{3}right)=2 ) and ( alpha beta=4 / 3 )
Let ( x=frac{alpha}{beta}+frac{beta}{alpha}+2left(frac{1}{alpha}+frac{1}{beta}right)+3 alpha beta )
( =frac{alpha^{2}+beta^{2}}{alpha beta}+2left(frac{alpha+beta}{alpha beta}right)+3 alpha beta )
We know, ( alpha^{2}+beta^{2}=(alpha+beta)^{2}-2 alpha beta=2^{2}-2 times frac{4}{3}=4-frac{8}{3}=frac{4}{3} )
( alpha^{2}+beta^{2}=frac{4}{3}, alpha+beta=2 ) and ( alpha beta=4 / 3 . ) So, ( x=frac{4 / 3}{4 / 3}+ )

# If a and B are the zeroes of the quadratic polynomial p(s) = 3s2 - 6s + 4, the value of α, β + β α ,71 1) + 2 -+- α β) + 3αβ is

Solution