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

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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))
$$ \\ $$$$\:{Q}:\:\:\:\:\alpha\:,\:\beta\:,\gamma\:{are}\:{the}\:{roots}\:{of}\:{the}\:{following} \\ $$$$\:\:\:\:\:{equation}\:.\:{find}\:{the}\:{value}\:{of}: \\ $$$$ \\ $$$$\:\:\:\:\:{Eq}^{\:{n}} \::\:\:\:{x}^{\:\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:{E}\:=\:\frac{\alpha}{\beta\:+\gamma}\:+\frac{\beta}{\alpha\:+\gamma}\:+\frac{\gamma}{\alpha+\:\beta} \\ $$$$ \\ $$
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)
$$\:\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} +\mathrm{x}+\mathrm{2}\:=\:\mathrm{0}\:\begin{cases}{\alpha}\\{\beta}\\{\gamma}\end{cases}\: \\ $$$$\:\mathrm{then}\:\mathrm{4x}^{\mathrm{3}} −\mathrm{5x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{1}=\mathrm{0}\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}−\alpha}}\\{\frac{\mathrm{1}}{\mathrm{2}−\beta}}\\{\frac{\mathrm{1}}{\mathrm{2}−\gamma}}\end{cases}\:\:\:\:\:\: \\ $$$$\:\mathrm{by}\:\mathrm{Vieta}'\mathrm{s}\: \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}−\alpha}\:+\:\frac{\mathrm{1}}{\mathrm{2}−\beta}\:+\:\frac{\mathrm{1}}{\mathrm{2}−\gamma}\:=\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$
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)
$$\:\mathrm{E}=\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}−\alpha}\:+\frac{\mathrm{1}}{\mathrm{2}−\beta}\:+\frac{\mathrm{1}}{\mathrm{2}−\gamma}\:\right)−\mathrm{3} \\ $$$$\:\mathrm{E}=\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)−\mathrm{3}=\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{3}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
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) ✓
$$\:\alpha,\beta,\gamma\:{are}\:{roots}\:{of}\:\:{x}^{\:\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{2}=\mathrm{0} \\ $$$${E}\:=\:\frac{\alpha}{\beta\:+\gamma}\:+\frac{\beta}{\alpha\:+\gamma}\:+\frac{\gamma}{\alpha+\:\beta}=? \\ $$$$\: \\ $$$$\alpha+\beta+\gamma=−\left(−\mathrm{2}\right)=\mathrm{2} \\ $$$$\alpha\beta+\beta\gamma+\gamma\alpha=\mathrm{1} \\ $$$$\alpha\beta\gamma=−\mathrm{2} \\ $$$${E}+\mathrm{3}=\:\frac{\alpha}{\beta\:+\gamma}+\mathrm{1}\:+\frac{\beta}{\alpha\:+\gamma}+\mathrm{1}\:+\frac{\gamma}{\alpha+\:\beta}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\alpha+\beta\:+\gamma}{\beta\:+\gamma}\:+\frac{\beta+\alpha\:+\gamma}{\alpha\:+\gamma}\:+\frac{\gamma+\alpha+\:\beta}{\alpha+\:\beta} \\ $$$$\:\:\:\:=\left(\alpha+\beta\:+\gamma\right)\left(\frac{\mathrm{1}}{\beta\:+\gamma}\:+\frac{\mathrm{1}}{\alpha\:+\gamma}\:+\frac{\mathrm{1}}{\alpha+\:\beta}\right) \\ $$$$\:=\left(\alpha+\beta\:+\gamma\right)\left(\frac{\mathrm{1}}{\alpha+\beta\:+\gamma−\alpha}\:+\frac{\mathrm{1}}{\alpha+\beta\:+\gamma−\beta}\:+\frac{\mathrm{1}}{\alpha+\:\beta+\gamma−\gamma}\right) \\ $$$$\:=\left(\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{2}−\alpha}\:+\frac{\mathrm{1}}{\mathrm{2}−\beta}\:+\frac{\mathrm{1}}{\mathrm{2}−\gamma}\right) \\ $$$$\:\frac{{E}+\mathrm{3}}{\mathrm{2}}=\frac{\left(\mathrm{2}−\alpha\right)\left(\mathrm{2}−\beta\right)+\left(\mathrm{2}−\beta\right)\left(\mathrm{2}−\gamma\right)+\left(\mathrm{2}−\gamma\right)\left(\mathrm{2}−\alpha\right)}{\left(\mathrm{2}−\alpha\right)\left(\mathrm{2}−\beta\right)\left(\mathrm{2}−\gamma\right)} \\ $$$$\:\:\:\:\:\:\:=\frac{−\mathrm{4}\left(\alpha+\beta+\gamma\right)+\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)+\mathrm{12}}{\mathrm{8}−\alpha\beta\gamma−\mathrm{4}\left(\alpha+\beta+\gamma\right)+\mathrm{2}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)} \\ $$$$\:\:\:\:\:\:\:=\frac{−\mathrm{4}\left(\mathrm{2}\right)+\left(\mathrm{1}\right)+\mathrm{12}}{\mathrm{8}−\left(−\mathrm{2}\right)−\mathrm{4}\left(\mathrm{2}\right)+\mathrm{2}\left(\mathrm{1}\right)}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${E}=\frac{\mathrm{5}}{\mathrm{4}}×\mathrm{2}−\mathrm{3}=−\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 30/Oct/23
thanks alot sir?
$${thanks}\:\:{alot}\:{sir}? \\ $$

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