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Question Number 145443 by 7770 last updated on 04/Jul/21
The roots of the equation 2x^2 +px+q=0 are 2α+β and α+2β. Find the values of p and q
$${The}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\:\:{are}\:\mathrm{2}\alpha+\beta\:\:{and} \\ $$$$\alpha+\mathrm{2}\beta.\:{Find}\:{the}\:{values}\:{of}\:{p}\:{and}\:{q} \\ $$
Answered by Olaf_Thorendsen last updated on 05/Jul/21
S = x_1 +x_2 = (2α+β)+(α+2β) = 3(α+β) P = x_1 x_2 = (2α+β)(α+2β) = 2α^2 +5αβ+2β^2 S = −(b/a) = −(p/2) = 3(α+β) ⇒ p = −6(α+β) P = (c/a) = (q/2) = 2α^2 +5αβ+2β^2 ⇒ q = 4α^2 +10αβ+4β^2
$$\mathrm{S}\:=\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} \:=\:\left(\mathrm{2}\alpha+\beta\right)+\left(\alpha+\mathrm{2}\beta\right)\:=\:\mathrm{3}\left(\alpha+\beta\right) \\ $$$$\mathrm{P}\:=\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} \:=\:\left(\mathrm{2}\alpha+\beta\right)\left(\alpha+\mathrm{2}\beta\right)\:=\:\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{5}\alpha\beta+\mathrm{2}\beta^{\mathrm{2}} \\ $$$$\mathrm{S}\:=\:−\frac{{b}}{{a}}\:=\:−\frac{{p}}{\mathrm{2}}\:=\:\mathrm{3}\left(\alpha+\beta\right) \\ $$$$\Rightarrow\:{p}\:=\:−\mathrm{6}\left(\alpha+\beta\right) \\ $$$$\mathrm{P}\:=\:\frac{{c}}{{a}}\:=\:\frac{{q}}{\mathrm{2}}\:=\:\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{5}\alpha\beta+\mathrm{2}\beta^{\mathrm{2}} \\ $$$$\Rightarrow\:{q}\:=\:\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{10}\alpha\beta+\mathrm{4}\beta^{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 05/Jul/21
2x^2 +px+q=0; 2α+β &α+2β are roots._(−) 2(2α+β)^2 +p(2α+β)+q....(i) 2(α+2β)^2 +p(α+2β)+q....(ii) (i)−(ii): 2{(2α+β)^2 −(α+2β)^2 }+p{(2α+β)−(α+2β)}=0 −6α^2 +6β^2 −p(α−β)=0 p=((−6α^2 +6β^2 )/(α−β)))=((−6(α^2 −β^2 ))/(α−β))=−6(α+β) p=−6(α+β) (i)⇒ q=−2(2α+β)^2 −p(2α+β) =−2(2α+β)^2 −{−6(α+β)}(2α+β) =−8α^2 −8αβ−2β^2 +6(2α^2 +3αβ+β^2 ) q=4α^2 +10αβ+4β^2
$$\underset{−} {\mathrm{2}{x}^{\mathrm{2}} +{px}+{q}=\mathrm{0};\:\mathrm{2}\alpha+\beta\:\&\alpha+\mathrm{2}\beta\:{are}\:{roots}.} \\ $$$$\:\:\:\:\:\:\mathrm{2}\left(\mathrm{2}\alpha+\beta\right)^{\mathrm{2}} +{p}\left(\mathrm{2}\alpha+\beta\right)+{q}….\left({i}\right) \\ $$$$\:\:\:\:\:\:\mathrm{2}\left(\alpha+\mathrm{2}\beta\right)^{\mathrm{2}} +{p}\left(\alpha+\mathrm{2}\beta\right)+{q}….\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\:\:\:\:\mathrm{2}\left\{\left(\mathrm{2}\alpha+\beta\right)^{\mathrm{2}} −\left(\alpha+\mathrm{2}\beta\right)^{\mathrm{2}} \right\}+{p}\left\{\left(\mathrm{2}\alpha+\beta\right)−\left(\alpha+\mathrm{2}\beta\right)\right\}=\mathrm{0} \\ $$$$−\mathrm{6}\alpha^{\mathrm{2}} +\mathrm{6}\beta^{\mathrm{2}} −{p}\left(\alpha−\beta\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{p}=\frac{−\mathrm{6}\alpha^{\mathrm{2}} +\mathrm{6}\beta^{\mathrm{2}} }{\left.\alpha−\beta\right)}=\frac{−\mathrm{6}\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)}{\alpha−\beta}=−\mathrm{6}\left(\alpha+\beta\right) \\ $$$$\:\:\:\:{p}=−\mathrm{6}\left(\alpha+\beta\right) \\ $$$$\left({i}\right)\Rightarrow\:{q}=−\mathrm{2}\left(\mathrm{2}\alpha+\beta\right)^{\mathrm{2}} −{p}\left(\mathrm{2}\alpha+\beta\right) \\ $$$$\:\:=−\mathrm{2}\left(\mathrm{2}\alpha+\beta\right)^{\mathrm{2}} −\left\{−\mathrm{6}\left(\alpha+\beta\right)\right\}\left(\mathrm{2}\alpha+\beta\right) \\ $$$$\:\:=−\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{8}\alpha\beta−\mathrm{2}\beta^{\mathrm{2}} +\mathrm{6}\left(\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{3}\alpha\beta+\beta^{\mathrm{2}} \right) \\ $$$$\:\:\:{q}=\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{10}\alpha\beta+\mathrm{4}\beta^{\mathrm{2}} \\ $$

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