Question

if α, β, γ are the roots of the equation x^3- 3x + 11 = 0 then find the equation whose roots are (α+β), (β+ γ) and (γ+ α).​

Answer 1

Answer:

x² - 6x + 9

Step-by-step explanation:

If alpha beta gamma are the roots of the equation x2 - 3x + 11 then find the equation whose roots are (alpha + beta)

α & β are roots of x² - 3x + 11

=> α + β = - (-3)/1 = 3    (Sum of roots)

α + β  is roots of other equation

let say γ is roots of other equation

so  γ  =  α + β  = 3

Sum of roots = γ + γ = 3 + 3 = 6

Products of roots = γ * γ = 3 * 3 = 9

Equation

x² - 6x + 9

or Simply

(x - γ)(x - γ)

= (x-3)(x-3)

= (x -3)²

= x² - 6x + 9

Answer 2

Answer:

Step-by-step explanation:

If α, β, γ are the roots of the equation x^3- 3x + 11 then-:

x³3- 3x + 11 = (x-α)(x-β)(x-γ)

Using the relation between the zeros-:

α+β+γ = -b/a = 0/1 = 0

=>α+β= -γ

=>β+γ= -α

=>γ+α= -β

αβγ = -d/a= -11

αβ+βγ+αγ = c/a = -3

The roots of our required equation are  (α+β),(β+γ)  and  (γ+α)  or rather  −α,−β  and  −γ

Our required equation is,

=>[x−(α+β)][x−(β+γ)][x−(γ+α)]=0  

∴(x+α)(x+β)(x+γ)=0  

∴x³+γx²+βx²+βγx+αx²+γαx+αβx+αβγ=0  

∴x²+(α+ β+ γ)x²+(αβ+βγ+γα)x+αβγ=0  

∴x3−3x−11=0