ROOTS OF POLYNOMIAL OF DEGREE 4

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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