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Question Number 220164 by mehdee7396 last updated on 06/May/25
if α^2 −5α+2=0 & β^2 −5β+2=0 then ((4α+β^5 )/(5β^2 ))=?
$${if}\:\:\alpha^{\mathrm{2}} −\mathrm{5}\alpha+\mathrm{2}=\mathrm{0}\:\:\&\:\:\beta^{\mathrm{2}} −\mathrm{5}\beta+\mathrm{2}=\mathrm{0} \\ $$$${then}\:\:\frac{\mathrm{4}\alpha+\beta^{\mathrm{5}} }{\mathrm{5}\beta^{\mathrm{2}} }=? \\ $$
Answered by Rasheed.Sindhi last updated on 07/May/25
if α^2 −5α+2=0 ∧ β^2 −5β+2=0 then ((4α+β^5 )/(5β^2 ))=? α^2 −5α+2=0 ...i ∧ β^2 −5β+2=0...ii ⇒x^2 −5x+2=0 where x=α , β α+β=5 ∧ αβ=2 ii⇒β^2 =5β−2 ((4α+β^5 )/(5β^2 ))=((4α+(5β−2)^2 β)/(5(5β−2)))=((4α+(25β^2 −20β+4)β)/(25β−10)) =((4α+(25(5β−2)−20β+4)β)/(25β−10)) =((4α+(105β−46)β)/(25β−10)) =((4α+105β^2 −46β)/(25β−10)) =((4α+105(5β−2)−46β)/(25β−10)) =((4α+525β−210−46β)/(25β−10)) =((4α+479β−210)/(25β−10)) =((4α+4β+475β−210)/(25β−10)) =((4(α+β)+475β−210)/(25β−10)) =((4(5)+475β−210)/(25β−10)) =((19(25β−10))/(25β−10))=19
$$ \\ $$$$\mathrm{if}\:\:\alpha^{\mathrm{2}} \:−\mathrm{5}\alpha+\mathrm{2}=\mathrm{0}\:\wedge\:\:\beta^{\mathrm{2}} \:−\mathrm{5}\beta+\mathrm{2}=\mathrm{0}\:\mathrm{then}\: \\ $$$$\:\frac{\mathrm{4}\alpha+\beta^{\mathrm{5}} }{\mathrm{5}\beta^{\mathrm{2}} }=? \\ $$$$\: \\ $$$$\alpha^{\mathrm{2}} \:−\mathrm{5}\alpha+\mathrm{2}=\mathrm{0}\:…{i}\:\wedge\:\:\beta^{\mathrm{2}} \:−\mathrm{5}\beta+\mathrm{2}=\mathrm{0}…{ii} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2}=\mathrm{0}\:{where}\:{x}=\alpha\:,\:\beta \\ $$$$\alpha+\beta=\mathrm{5}\:\wedge\:\alpha\beta=\mathrm{2} \\ $$$${ii}\Rightarrow\beta^{\mathrm{2}} =\mathrm{5}\beta−\mathrm{2} \\ $$$$\:\frac{\mathrm{4}\alpha+\beta^{\mathrm{5}} }{\mathrm{5}\beta^{\mathrm{2}} }=\frac{\mathrm{4}\alpha+\left(\mathrm{5}\beta−\mathrm{2}\right)^{\mathrm{2}} \beta}{\mathrm{5}\left(\mathrm{5}\beta−\mathrm{2}\right)}=\frac{\mathrm{4}\alpha+\left(\mathrm{25}\beta^{\mathrm{2}} −\mathrm{20}\beta+\mathrm{4}\right)\beta}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\alpha+\left(\mathrm{25}\left(\mathrm{5}\beta−\mathrm{2}\right)−\mathrm{20}\beta+\mathrm{4}\right)\beta}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\alpha+\left(\mathrm{105}\beta−\mathrm{46}\right)\beta}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\alpha+\mathrm{105}\beta^{\mathrm{2}} −\mathrm{46}\beta}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\alpha+\mathrm{105}\left(\mathrm{5}\beta−\mathrm{2}\right)−\mathrm{46}\beta}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\alpha+\mathrm{525}\beta−\mathrm{210}−\mathrm{46}\beta}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\alpha+\mathrm{479}\beta−\mathrm{210}}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\alpha+\mathrm{4}\beta+\mathrm{475}\beta−\mathrm{210}}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\left(\alpha+\beta\right)+\mathrm{475}\beta−\mathrm{210}}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{4}\left(\mathrm{5}\right)+\mathrm{475}\beta−\mathrm{210}}{\mathrm{25}\beta−\mathrm{10}} \\ $$$$=\frac{\mathrm{19}\left(\mathrm{25}\beta−\mathrm{10}\right)}{\mathrm{25}\beta−\mathrm{10}}=\mathrm{19} \\ $$
Commented by mehdee7396 last updated on 07/May/25
good
$${good} \\ $$
Answered by mnjuly1970 last updated on 07/May/25
((4α^3 +α^2 β^5 )/(5α^2 β^2 ))=((4α^3 +α^2 β^2 β^3 )/(20)) , αβ=2 , α+β=5 =((4(α^3 +β^3 ))/(5(αβ)^2 ))=((4((α+β)^3 −3αβ(α+β)))/(20))= ((4(5^3 −3(2)(5)))/(20)) =((4(95))/(20))=((95)/5)=19
$$\:\:\:\:\:\:\:\frac{\mathrm{4}\alpha^{\mathrm{3}} +\alpha^{\mathrm{2}} \beta^{\mathrm{5}} }{\mathrm{5}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} }=\frac{\mathrm{4}\alpha^{\mathrm{3}} +\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \beta^{\mathrm{3}} }{\mathrm{20}}\:\:\:\:,\:\alpha\beta=\mathrm{2}\:,\:\alpha+\beta=\mathrm{5} \\ $$$$\:\:\:\:=\frac{\mathrm{4}\left(\alpha^{\mathrm{3}} \:+\beta^{\mathrm{3}} \right)}{\mathrm{5}\left(\alpha\beta\right)^{\mathrm{2}} }=\frac{\mathrm{4}\left(\left(\alpha+\beta\right)^{\mathrm{3}} −\mathrm{3}\alpha\beta\left(\alpha+\beta\right)\right)}{\mathrm{20}}=\:\frac{\mathrm{4}\left(\mathrm{5}^{\mathrm{3}} −\mathrm{3}\left(\mathrm{2}\right)\left(\mathrm{5}\right)\right)}{\mathrm{20}} \\ $$$$\:=\frac{\mathrm{4}\left(\mathrm{95}\right)}{\mathrm{20}}=\frac{\mathrm{95}}{\mathrm{5}}=\mathrm{19} \\ $$
Commented by mehdee7396 last updated on 07/May/25
very nice ⋛
$${very}\:\:\:{nice}\:\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by mnjuly1970 last updated on 07/May/25
⋛
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$

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