Let ax^{4}+bx^{3}+cx^{2}+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

x^{4}+2x^{3}-25x^{2}-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)

= (x^{2}-(α+β)x+αβ)(x^{2}-(γ+δ)x+γδ)

x^{4}+2x^{3}-25x^{2}-26x+120 = x^{4} - x^{3}[(γ+δ)+(α+β)]+[γδ+αβ+(α+β)(γ+δ)]x^{2}-[αβ(γ+δ)+γδ(α+β)]x+αβγδ

Equating the coefficients of x^{3}, x^{2}, 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) x^{2}-(α+β)x + αβ

(ii) x^{2}-(γ+δ)x+γδ

By solving x^{2}-6x+8, we will get values of the roots α and β

x^{2}-6x+8 = 0

(x-4)(x-2) = 0

x = 4 and x = 2

By solving x^{2}-8x+15, we will get values of the roots γ and δ

x^{2}-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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