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f-x-x-4-2x-3-3x-2-4x-5-Find-the-roots-

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Question Number 39779 by ajfour last updated on 10/Jul/18
f(x)=x^4 −2x^3 +3x^2 −4x+5 Find the roots.
$${f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5} \\ $$$${Find}\:{the}\:{roots}. \\ $$
Answered by MJS last updated on 11/Jul/18
approximate: x_1 =−.287815−1.41609i x_2 =−.287815+1.41609i x_3 =1.28782−.857897i x_4 =1.28782+.857897i exact: the usual procedure, but we have two pairs of conjugated complex solutions f(x)=(x−α−(√β)i)(x−α+(√β)i)(x−γ−(√δ)i)(x−γ+(√δ)i) ⇒ α=1−γ β=((γ(2γ^2 −3γ+3))/(2γ−1)) δ=(((γ−1)(2γ^2 −γ+2))/(2γ−1)) γ^6 −3γ^5 +(9/2)γ^4 −4γ^3 +((21)/(16))γ^2 +(3/(16))γ−(3/(16))=0 γ=u+(1/2) u^6 +(3/4)u^4 −(3/4)u^2 −(1/(16))=0 u=(√v) v^3 +(3/4)v^2 −(3/4)v−(1/(16))=0 v=w−(1/4) w^3 −((15)/(16))w+(5/(32))=0 this has got 3 real solutions, so we need the trigonometric method. w_k =2(√(−(p/3)))sin((1/3)(arcsin(((9q)/(2p^2 ))(√(−(p/3))))+2kπ)) with k=0, 1, 2 p=−((15)/(16)); q=(5/(32)) w_0 =((√5)/2)sin((1/3)arcsin ((√5)/5)) w_1 =((√5)/2)cos((π/6)+(1/3)arcsin ((√5)/5)) w_2 =−((√5)/2)sin((π/3)+(1/3)arcsin ((√5)/5)) again the way back is hard if you want exact solutions. all three w_k lead to the same set of α, β, γ, δ but I recommend to take w_1 to get real values in each step back.
$$\mathrm{approximate}: \\ $$$${x}_{\mathrm{1}} =−.\mathrm{287815}−\mathrm{1}.\mathrm{41609i} \\ $$$${x}_{\mathrm{2}} =−.\mathrm{287815}+\mathrm{1}.\mathrm{41609i} \\ $$$${x}_{\mathrm{3}} =\mathrm{1}.\mathrm{28782}−.\mathrm{857897i} \\ $$$${x}_{\mathrm{4}} =\mathrm{1}.\mathrm{28782}+.\mathrm{857897i} \\ $$$$\mathrm{exact}: \\ $$$$\mathrm{the}\:\mathrm{usual}\:\mathrm{procedure},\:\mathrm{but}\:\mathrm{we}\:\mathrm{have}\:\mathrm{two}\:\mathrm{pairs} \\ $$$$\mathrm{of}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$${f}\left({x}\right)=\left({x}−\alpha−\sqrt{\beta}\mathrm{i}\right)\left({x}−\alpha+\sqrt{\beta}\mathrm{i}\right)\left({x}−\gamma−\sqrt{\delta}\mathrm{i}\right)\left({x}−\gamma+\sqrt{\delta}\mathrm{i}\right) \\ $$$$\Rightarrow\:\alpha=\mathrm{1}−\gamma \\ $$$$\:\:\:\:\:\:\beta=\frac{\gamma\left(\mathrm{2}\gamma^{\mathrm{2}} −\mathrm{3}\gamma+\mathrm{3}\right)}{\mathrm{2}\gamma−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\delta=\frac{\left(\gamma−\mathrm{1}\right)\left(\mathrm{2}\gamma^{\mathrm{2}} −\gamma+\mathrm{2}\right)}{\mathrm{2}\gamma−\mathrm{1}} \\ $$$$\gamma^{\mathrm{6}} −\mathrm{3}\gamma^{\mathrm{5}} +\frac{\mathrm{9}}{\mathrm{2}}\gamma^{\mathrm{4}} −\mathrm{4}\gamma^{\mathrm{3}} +\frac{\mathrm{21}}{\mathrm{16}}\gamma^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{16}}\gamma−\frac{\mathrm{3}}{\mathrm{16}}=\mathrm{0} \\ $$$$\gamma={u}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${u}^{\mathrm{6}} +\frac{\mathrm{3}}{\mathrm{4}}{u}^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{4}}{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$${u}=\sqrt{{v}} \\ $$$${v}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{4}}{v}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{4}}{v}−\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$${v}={w}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${w}^{\mathrm{3}} −\frac{\mathrm{15}}{\mathrm{16}}{w}+\frac{\mathrm{5}}{\mathrm{32}}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{got}\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions},\:\mathrm{so}\:\mathrm{we}\:\mathrm{need}\:\mathrm{the} \\ $$$$\mathrm{trigonometric}\:\mathrm{method}. \\ $$$${w}_{{k}} =\mathrm{2}\sqrt{−\frac{{p}}{\mathrm{3}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{arcsin}\left(\frac{\mathrm{9}{q}}{\mathrm{2}{p}^{\mathrm{2}} }\sqrt{−\frac{{p}}{\mathrm{3}}}\right)+\mathrm{2}{k}\pi\right)\right)\:\mathrm{with}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$${p}=−\frac{\mathrm{15}}{\mathrm{16}};\:{q}=\frac{\mathrm{5}}{\mathrm{32}} \\ $$$${w}_{\mathrm{0}} =\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\right) \\ $$$${w}_{\mathrm{1}} =\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\right) \\ $$$${w}_{\mathrm{2}} =−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\right) \\ $$$$\mathrm{again}\:\mathrm{the}\:\mathrm{way}\:\mathrm{back}\:\mathrm{is}\:\mathrm{hard}\:\mathrm{if}\:\mathrm{you}\:\mathrm{want} \\ $$$$\mathrm{exact}\:\mathrm{solutions}.\:\mathrm{all}\:\mathrm{three}\:{w}_{{k}} \:\mathrm{lead}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{set}\:\mathrm{of}\:\alpha,\:\beta,\:\gamma,\:\delta\:\mathrm{but}\:\mathrm{I}\:\mathrm{recommend}\:\mathrm{to} \\ $$$$\mathrm{take}\:{w}_{\mathrm{1}} \:\mathrm{to}\:\mathrm{get}\:\mathrm{real}\:\mathrm{values}\:\mathrm{in}\:\mathrm{each}\:\mathrm{step}\:\mathrm{back}. \\ $$
Commented by MJS last updated on 10/Jul/18
Ferrari is very complicated sometimes but yes, it always gives all solutions. Still the problem witb the 3 real solutions occurs...
$$\mathrm{Ferrari}\:\mathrm{is}\:\mathrm{very}\:\mathrm{complicated}\:\mathrm{sometimes}\:\mathrm{but} \\ $$$$\mathrm{yes},\:\mathrm{it}\:\mathrm{always}\:\mathrm{gives}\:\mathrm{all}\:\mathrm{solutions}.\:\mathrm{Still}\:\mathrm{the} \\ $$$$\mathrm{problem}\:\mathrm{witb}\:\mathrm{the}\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{occurs}… \\ $$
Commented by ajfour last updated on 10/Jul/18
I read a little of Ferrari′s solution.
$${I}\:{read}\:{a}\:{little}\:{of}\:{Ferrari}'{s}\:{solution}. \\ $$
Commented by ajfour last updated on 11/Jul/18
Thanks for this much,Sir. Whats the trigonometric way for our cubic eq. ?
$${Thanks}\:{for}\:{this}\:{much},{Sir}. \\ $$$${Whats}\:{the}\:{trigonometric}\:{way} \\ $$$${for}\:{our}\:{cubic}\:{eq}.\:? \\ $$

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