Let ax4+bx3+cx2+dx+e be the polynomial of degree 4 whose roots are α, β, γ and δ
Formula :
α + β + γ + δ = - b (co-efficient of x³)
α β + β γ + γ δ + δ α = c (co-efficient of x²)
α β γ + β γ δ + γ δ α + δ α β = - d (co-efficient of x)
α β γ δ = e
Example :
Solve the equation
x4+2x3-25x2-26x+120 = 0
given that the product of two roots is 8.
Solution :
Let α, β, γ, δ be the four roots.
Product of two roots = 8
αβ = 8
= (x - α)(x - β)(x - γ)(x - δ)
= (x2-(α+β)x+αβ)(x2-(γ+δ)x+γδ)
x4+2x3-25x2-26x+120 = x4 - x3[(γ+δ)+(α+β)]+[γδ+αβ+(α+β)(γ+δ)]x2-[αβ(γ+δ)+γδ(α+β)]x+αβγδ
Equating the coefficients of x3, x2, x and constant
-[(γ+δ)+(α+β)] = 2
α+β+γ+δ = -2 --------------- (1)
γδ+αβ+(α+β)(γ+δ) = -25 -----------(2)
-[αβ(γ+δ) + γδ(α+β)] = -26 ------------(3)
αβγδ = 120 ---------(4)
8γδ = 120
γδ = 120/8
γδ = 15
By applying αβ = 8 and γδ = 15 in (3), we get
8(γ+δ)+15(α+β) = 26 ----------(5)
Multiplying the first equation by 8
8(α+β)+8(γ+δ) = -16 -------------(6)
(5) - (6)
7(α+β) = 42
(α+β) = 6
By applying α+β = 6 in (5), we get
15(6)+8(γ+δ) = 26
90+8(γ+δ) = 26
8(γ+δ) = -64
(γ+δ) = -8
By applying αβ = 8, α+β = 6, γ+δ = -8 in (2)
γδ+αβ+(α+β)(γ+δ) = -25
γδ+8+(6)(-8) = -25
γδ+8-48 = -25
γδ-40 = -25
γδ = 15
Now let us take the two quadratic equations
(i) x2-(α+β)x + αβ
(ii) x2-(γ+δ)x+γδ
By solving x2-6x+8, we will get values of the roots α and β
x2-6x+8 = 0
(x-4)(x-2) = 0
x = 4 and x = 2
By solving x2-8x+15, we will get values of the roots γ and δ
x2-8x+15 = 0
(x-3)(x-5) = 0
x = 3 and x = 5
Therefore the four roots are 2, 3, 4, 5.
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