If alpha and beta are zeroes of polynomial 3x2 + 2x - 6 find the value of alpha-beta, alpha2+ beta2, alpha3+ beta3, and 1/alpha + 1/beta. Share with your friends Share 12 Manbar Singh answered this The given polynomial is:3x2 + 2x - 6Comparing this polynomial with standard polynomial ax2 + bx + c, we get,a = 3 , b = 2 and c =-6Since, α & β are the zeroes of the given polynomial 3x2 + 2x - 6, then,α + β = -ba = -23And αβ = ca = -63 = -21We know that α - β2 = α + β2 - 4 αβ⇒ α - β2 = -232 - 4 × -2⇒α - β2 = 49 + 8⇒α - β2 = 4 + 729 = 769Therefore, α - β = 763 = 4 × 193 = 2193 2We know that α + β2 = α2 + β2 + 2 αβ⇒ -232 = α2 + β2 + 2 -2⇒ 49 = α2 + β2 - 4⇒ 49 + 4 = α2 + β2⇒ 4 + 369 = α2 + β2⇒ α2 + β2 = 409 3We know that α + β3 = α3 + β3 + 3 αβ α + β⇒ -233 = α3 + β3 + 3 -2 -23⇒ -827 = α3 + β3 + 4⇒ -827 - 4 = α3 + β3⇒ -8 - 10827 = α3 + β3⇒ α3 + β3 = -11627 4 1α + 1β= β + ααβ = α + βαβ = -23-21 = -23 × -12 = 13 32 View Full Answer
