Menu Close

Q-are-the-roots-of-the-following-equation-find-the-value-of-Eq-n-x-3-2x-2-x-2-0-E-




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:x32x2+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)
x32x2+x+2=0{αβγthen4x35x2+4x1=0{12α12β12γbyVietas12α+12β+12γ=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(12α+12β+12γ)3E=2(54)3=523=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,γarerootsofx32x2+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+γα+1α+β+γβ+1α+β+γγ)=(2)(12α+12β+12γ)E+32=(2α)(2β)+(2β)(2γ)+(2γ)(2α)(2α)(2β)(2γ)=4(α+β+γ)+(αβ+βγ+γα)+128αβγ4(α+β+γ)+2(αβ+βγ+γα)=4(2)+(1)+128(2)4(2)+2(1)=54E=54×23=12
Commented by mnjuly1970 last updated on 30/Oct/23
thanks  alot sir?
thanksalotsir?

Leave a Reply

Your email address will not be published. Required fields are marked *