Question

Alpha and beta are roots of this equation; 4_x²-7x+3=0 then find alpha cube + beta cube and also find 1 upon alpha square + 1 upon beta square

Answer 1

The given equation is 4x^2 - 7x+ 3= 0(i) (common equation is ax^2+bx+c)

According to the problem, α and β are the roots of the equation (i)

Therefore,

α + β = -(b/a) = -(-7/4)=7/4

and αβ =c/a=3/4
α^3 + β^3 = (α + β)^3 - 3αβ(α + β) = (7/4)^3 –3*(3/4)*(7/4)=(342/64)-(63/16)=(91/64)
1/(α^2)+1/(β^2)=(α^2 + β^2 )/(α^2 β^2 )..(ii)
Now α^2 + β^2 = (α + β)^2 - 2αβ =(7/4)^2-2(3/4)
=25/16
and α^2 β^2 =(αβ)^2=(3/4)^2=9/16
(ii) implies (25/16)÷(9/16)
=25/9