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Question Number 199109 by mnjuly1970 last updated on 28/Oct/23

Q: α , β ,γ are the roots of the following equation . find the value of: Eq^( n) : x^( 3) −2x^2 + x + 2=0 E = (α/(β +γ)) +(β/(α +γ)) +(γ/(α+ β))

Q:α,β,γaretherootsofthefollowingequation.findthevalueof:Eqn:x3−2x2+x+2=0E=αβ+γ+βα+γ+γα+β

Commented by cortano12 last updated on 28/Oct/23

x^3 −2x^2 +x+2 = 0 { (α),(β),(γ) :} then 4x^3 −5x^2 +4x−1=0 { ((1/(2−α))),((1/(2−β))),((1/(2−γ))) :} by Vieta′s (1/(2−α)) + (1/(2−β)) + (1/(2−γ)) = (5/4)

x3−2x2+x+2=0{αβγthen4x3−5x2+4x−1=0{12−α12−β12−γbyVieta′s12−α+12−β+12−γ=54

Answered by cortano12 last updated on 28/Oct/23

E= 2((1/(2−α)) +(1/(2−β)) +(1/(2−γ)) )−3 E=2((5/4))−3=(5/2)−3=−(1/2)

E=2(12−α+12−β+12−γ)−3E=2(54)−3=52−3=−12

Answered by Rasheed.Sindhi last updated on 28/Oct/23

α,β,γ are roots of x^( 3) −2x^2 + x + 2=0 E = (α/(β +γ)) +(β/(α +γ)) +(γ/(α+ β))=? α+β+γ=−(−2)=2 αβ+βγ+γα=1 αβγ=−2 E+3= (α/(β +γ))+1 +(β/(α +γ))+1 +(γ/(α+ β))+1 = ((α+β +γ)/(β +γ)) +((β+α +γ)/(α +γ)) +((γ+α+ β)/(α+ β)) =(α+β +γ)((1/(β +γ)) +(1/(α +γ)) +(1/(α+ β))) =(α+β +γ)((1/(α+β +γ−α)) +(1/(α+β +γ−β)) +(1/(α+ β+γ−γ))) =(2)((1/(2−α)) +(1/(2−β)) +(1/(2−γ))) ((E+3)/2)=(((2−α)(2−β)+(2−β)(2−γ)+(2−γ)(2−α))/((2−α)(2−β)(2−γ))) =((−4(α+β+γ)+(αβ+βγ+γα)+12)/(8−αβγ−4(α+β+γ)+2(αβ+βγ+γα))) =((−4(2)+(1)+12)/(8−(−2)−4(2)+2(1)))=(5/4) E=(5/4)×2−3=−(1/2) ✓

α,β,γarerootsofx3−2x2+x+2=0E=αβ+γ+βα+γ+γα+β=?α+β+γ=−(−2)=2αβ+βγ+γα=1αβγ=−2E+3=αβ+γ+1+βα+γ+1+γα+β+1=α+β+γβ+γ+β+α+γα+γ+γ+α+βα+β=(α+β+γ)(1β+γ+1α+γ+1α+β)=(α+β+γ)(1α+β+γ−α+1α+β+γ−β+1α+β+γ−γ)=(2)(12−α+12−β+12−γ)E+32=(2−α)(2−β)+(2−β)(2−γ)+(2−γ)(2−α)(2−α)(2−β)(2−γ)=−4(α+β+γ)+(αβ+βγ+γα)+128−αβγ−4(α+β+γ)+2(αβ+βγ+γα)=−4(2)+(1)+128−(−2)−4(2)+2(1)=54E=54×2−3=−12✓

Commented by mnjuly1970 last updated on 30/Oct/23

thanks alot sir?

thanksalotsir?

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