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If α, β, γ are the roots of the equation x3 - 3x + 11 =0 then find the equation whose roots are α + β, β+γ and γ+α.

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Solution

Dear Student,

x3-3x+11=0Comparing with ax3+bx2+cx+d=0Here, a=1b=0c=-3d=11We know that for a cubic equation, α+β+γ=-ba=-01=0 ...(i)and, αβ+βγ+γα = ca=-31=-3and, αβγ=-da=-111=-11Now, if there is a cubic polynomial of roots α+β, β+γ and γ+α, then it zeroes will be -γ, -α and -β respectively ( from eq i)Sum of zeroes=(-γ) + (-α) + (-β) = -(γ+α+β)=0Sum of product of its two zeroes=(-α)(-β)+(-β)(-γ)+(-γ)(-α)or, αβ+βγ+γα=-3and, Product of its zeroes=(-α)(-β)(-γ)=-αβγ=-(-11)=11So, required polynomial = x3-(Sum of zeroes)x2 +(Sum of product of its two zeroes)x-(Product of its zeroes) =x3-0x2-3x-11 =x3-3x-11

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