Quadratic Equations

Answer :

Let α & β are the roots of given equation.

                             α + β = – 8, α ∙ β = (3k / 2)

                   also (α/β) = (4/5)

                             ∴ α = 4m, β = 5m

                             ∴ α + β = 4m + 5m = – 8

                             ∴ m = {(– 8) / 9}

                             ∴ α = {(– 32) / 9} and β = {(– 40) / 9}

                   also αβ = (3k / 2) hence

                             {(– 32) / 9} × {(– 40) / 9} = (3/2) k

                             ∴ k = {(2560) / (243)}

Answer :

          x² – x + 1 = 0

                   x³ + 1 = (x² – x + 1) (x + 1) = 0

                   ∴ x³ = – 1

                   if α and β are roots then α + β = 1, αβ = 1

                   now x³ = – 1 hence α³ = – 1, β³ = – 1

                               α²⁰⁰⁹ + β²⁰⁰⁹         = {(α²⁰¹⁰) / α} + {(β²⁰¹⁰) / β}

                                                = [{(α³)⁶⁷⁰} / α] + [{(β³)⁶⁷⁰} / β]

                                                = [{(– 1)⁶⁷⁰} / α] + [{(– 1)⁶⁷⁰} / β]

                                                = (1/α) + (1/β)

                                                = {(α + β) / (αβ)} = (1/1) = 1

Answer :

ax² – 6x + c + 9 = 0                                                       (1)

                   one root is 3 – 5i hence other root has to be 3 + 5i

                             ∴ α = 3 – 5i

                             β = 3 + 5i

                             α + β = 6 and α × β = 9 + 25 = 34

                   from equation (1),

                                (6/a)         = α + β and {(c + 9) / a} = αβ

                             ∴ (6/a)        = 6 and [(c + 9) / a] = 34

                                  ∴ a        = 1 and c + 9 = 34

                                  ∴ a        = 1 and c = 25

Answer :

a(b – c)x² + b(c – a)x + c(a – b) = 0

                   roots are equal hence Δ = 0

                             ∴ b²(c – a)² – 4 ac (b – c)(a – b) = 0

                             [b (a + c) – 2ac]² = 0

                             ∴ ab + c – 2ac = 0
                             ∴ ab – ac = ac – bc

                             ∴ dividing by abc

                   (1/c) – (1/b) = (1/b) – (1/b)

                             ∴ (1/a), (1/b), (1/c) are in A.P.

                             ∴ a, b, c are in H.P

Answer :

bx² + cx + a = 0

                   roots are complex, Δ < 0

                             ∴ c² – 4 ab < 0

                             ∴ c² < 4 ab

                             ∴ – c² > – 4 ab                                                               (1)

                   given: 3b² x² + 6bcx + 2c²

                             = 3(bx + c²) – c²

                   Now 3(bx + c)² – c² > – c²

                   ∴ from (1)

                             3(bx + c)² – c² > – 4 ab