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Question Number 113211 by bemath last updated on 11/Sep/20
If α is a root of the equation 4x^2 +2x−1=0 . How do you prove the other root is 4α^3 −3α ?
$$\mathrm{If}\:\alpha\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{4x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{1}=\mathrm{0}\:.\:\mathrm{How}\:\mathrm{do}\:\mathrm{you} \\ $$$$\mathrm{prove}\:\mathrm{the}\:\mathrm{other}\:\mathrm{root}\:\mathrm{is} \\ $$$$\mathrm{4}\alpha^{\mathrm{3}} −\mathrm{3}\alpha\:?\: \\ $$
Answered by 1549442205PVT last updated on 11/Sep/20
α is a root of the equation 4x^2 +2x−1=0 (1) ,so we have 4α^2 +2α−1=0 ⇒4α^3 =α(1−2α)=α−2α^2 .From that 4α^3 −3α=−2α^2 −2α.Replace into(1) we get 4(2α^2 +2α)^2 −2(2α^2 +2α)−1 =16(α^4 +2α^3 +α^2 )−4α^2 −4α−1 =16α^4 +32α^3 +12α^2 −4α−1 =(4α^2 +2α−1)^2 +16α^3 +16α^2 −2 =16α^3 +16α^2 −2=4α(4α^2 +2α−1) +8α^2 +4α−2=8α^2 +4α−2 =2(4α^2 +2α−1)=0 This show that 4α^3 −3α is also other root of the equation 4x^2 +2x−1=0(q.e.d)
$$\alpha\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{4x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{1}=\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$,\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{2}\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\alpha^{\mathrm{3}} =\alpha\left(\mathrm{1}−\mathrm{2}\alpha\right)=\alpha−\mathrm{2}\alpha^{\mathrm{2}} .\mathrm{From}\:\mathrm{that} \\ $$$$\mathrm{4}\alpha^{\mathrm{3}} −\mathrm{3}\alpha=−\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{2}\alpha.\mathrm{Replace}\:\mathrm{into}\left(\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{4}\left(\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{2}\alpha\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{2}\alpha\right)−\mathrm{1} \\ $$$$=\mathrm{16}\left(\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{3}} +\alpha^{\mathrm{2}} \right)−\mathrm{4}\alpha^{\mathrm{2}} −\mathrm{4}\alpha−\mathrm{1} \\ $$$$=\mathrm{16}\alpha^{\mathrm{4}} +\mathrm{32}\alpha^{\mathrm{3}} +\mathrm{12}\alpha^{\mathrm{2}} −\mathrm{4}\alpha−\mathrm{1} \\ $$$$=\left(\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{2}\alpha−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{16}\alpha^{\mathrm{3}} +\mathrm{16}\alpha^{\mathrm{2}} −\mathrm{2} \\ $$$$=\mathrm{16}\alpha^{\mathrm{3}} +\mathrm{16}\alpha^{\mathrm{2}} −\mathrm{2}=\mathrm{4}\alpha\left(\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{2}\alpha−\mathrm{1}\right) \\ $$$$+\mathrm{8}\alpha^{\mathrm{2}} +\mathrm{4}\alpha−\mathrm{2}=\mathrm{8}\alpha^{\mathrm{2}} +\mathrm{4}\alpha−\mathrm{2} \\ $$$$=\mathrm{2}\left(\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{2}\alpha−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{This}\:\mathrm{show}\:\mathrm{that}\:\mathrm{4}\alpha^{\mathrm{3}} −\mathrm{3}\alpha\:\mathrm{is}\:\mathrm{also} \\ $$$$\mathrm{other}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{4x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{1}=\mathrm{0}\left(\mathrm{q}.\mathrm{e}.\mathrm{d}\right) \\ $$
Answered by $@y@m last updated on 11/Sep/20
4x^2 +2x−1=0 x=((−2±(√(4+16)))/8) x=((−2±(√(20)))/8) x=((−2±2(√5))/8) x=((−1±(√5))/4) Let α=((−1+(√5))/4), β=((−1−(√5))/4) Now, 4α^3 −3α=α(4α^2 −3) =((−1+(√5))/4){4(((−1+(√5))/4))^2 −3} =((−1+(√5))/4){((5+1−2(√5))/4)−3} =((−1+(√5))/4){((6−2(√5))/4)−3} =((−1+(√5))/4){((3−(√5))/2)−3} =((−1+(√5))/4){((−3−(√5))/2)} =(1/(32))(3+(√5)−5−3(√5)) =((−2−2(√5))/8) =((−1−(√5))/4) =β Q.E.D. Remark: You may assume β=((−1+(√5))/4), α=((−1−(√5))/4) and still you can show that 4α^3 −3α=β
$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{16}}}{\mathrm{8}} \\ $$$${x}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{20}}}{\mathrm{8}} \\ $$$${x}=\frac{−\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$${x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${Let}\:\alpha=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}},\:\beta=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${Now}, \\ $$$$\mathrm{4}\alpha^{\mathrm{3}} −\mathrm{3}\alpha=\alpha\left(\mathrm{4}\alpha^{\mathrm{2}} −\mathrm{3}\right) \\ $$$$=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\left\{\mathrm{4}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{3}\right\} \\ $$$$=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\left\{\frac{\mathrm{5}+\mathrm{1}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}−\mathrm{3}\right\} \\ $$$$=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\left\{\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}−\mathrm{3}\right\} \\ $$$$=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\left\{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{3}\right\} \\ $$$$=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\left\{\frac{−\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{3}+\sqrt{\mathrm{5}}−\mathrm{5}−\mathrm{3}\sqrt{\mathrm{5}}\right) \\ $$$$=\frac{−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$=\beta \\ $$$${Q}.{E}.{D}. \\ $$$${Remark}: \\ $$$${You}\:{may}\:{assume}\: \\ $$$$\beta=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}},\:\alpha=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${and}\:{still}\:{you}\:{can}\:{show}\:{that}\:\mathrm{4}\alpha^{\mathrm{3}} −\mathrm{3}\alpha=\beta \\ $$
Commented by 1549442205PVT last updated on 12/Sep/20
cos((2π)/5)=(((√5)−1)/4),((−1−(√5))/4)=cos((4π)/5) =cos(((4π)/5)−2π)=cos((−6π)/5)=cos((6π)/5) =4cos^3 ((2π)/5)−3cos((2π)/5).Hence,put α=cos((2π)/5) =(((√5)−1)/4) then 4α^3 −3α=((−1−(√5))/4)
$$\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{5}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}},\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}=\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{5}} \\ $$$$=\mathrm{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}−\mathrm{2}\pi\right)=\mathrm{cos}\frac{−\mathrm{6}\pi}{\mathrm{5}}=\mathrm{cos}\frac{\mathrm{6}\pi}{\mathrm{5}} \\ $$$$=\mathrm{4cos}^{\mathrm{3}} \frac{\mathrm{2}\pi}{\mathrm{5}}−\mathrm{3cos}\frac{\mathrm{2}\pi}{\mathrm{5}}.\mathrm{Hence},\mathrm{put}\:\alpha=\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{5}} \\ $$$$=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\:\mathrm{then}\:\mathrm{4}\alpha^{\mathrm{3}} −\mathrm{3}\alpha=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$
Answered by Dwaipayan Shikari last updated on 11/Sep/20
α+β=−(1/2) αβ=−(1/4) (α+β)^2 −4αβ=(α−β)^2 (1/4)+1=(α−β)^2 ⇒(α−β)=±((√5)/2) α=(((√5)−1)/4) or−((((√5)+1)/4)) β=((−(√5)−1)/4) or ((((√5)−1)/4)) 4α^3 −3α=−4((((√5)+1)/4))^3 +3((((√5)+1)/4))=(((√5)−1)/4)=β 4α^3 −3α=4((((√5)−1)/4))^3 −3((((√5)−1)/4))=β
$$\alpha+\beta=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\alpha\beta=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta=\left(\alpha−\beta\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}=\left(\alpha−\beta\right)^{\mathrm{2}} \Rightarrow\left(\alpha−\beta\right)=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\alpha=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\:{or}−\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\beta=\frac{−\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\:{or}\:\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\mathrm{4}\alpha^{\mathrm{3}} −\mathrm{3}\alpha=−\mathrm{4}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} +\mathrm{3}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}=\beta \\ $$$$\mathrm{4}\alpha^{\mathrm{3}} −\mathrm{3}\alpha=\mathrm{4}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)=\beta \\ $$

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