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Question Number 36207 by MJS last updated on 30/May/18
x^4 +10x^3 +6x−1=0 for those who want an exact solution... (x−a−bi)(x−a+bi)(x−c−d)(x−c+d)=0 x^4 −2(a+c)x^3 +(a^2 +4ac+b^2 +c^2 −d^2 )x^2 − −2(a(c^2 −d^2 )+c(a^2 +b^2 ))x+(a^2 +b^2 )(c^2 −d^2 )=0 1. −2(a+c)=10 2. a^2 +4ac+b^2 +c^2 −d^2 =0 3. −2(a(c^2 −d^2 )+c(a^2 +b^2 ))=6 4. (a^2 +b^2 )(c^2 −d^2 )=−1 1. a=−c−5e 2. b=(√(−a^2 +4ac−c^2 +d^2 ))=(√(2c^2 +10c+d^2 −25)) 3. d=(√(ac+c^2 +((b^2 c+3)/a)))=(√(−((2c^3 +15c^2 +3)/(2c+5)))) 2. b=(√((2c^3 +15c^2 −128)/(2c+5))) 4. 4c^4 +40c^3 +75c^2 −125c+((689)/4)−((17161)/(4(2c+5)^2 ))=0 64(c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1)=0 c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1=0 c=u−((15)/6) u^6 −((75)/4)u^4 +((1939)/(16))u^2 −((17161)/(64))=0 u=(√v) v^3 −((75)/4)v^2 +((1939)/(16))v−((17161)/(64))=0 v=w+((75)/(12)) w^3 +4w+1=0 w=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) +((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) = =((−(1/2)+((√(849))/(18))))^(1/3) +((−(1/2)−((√(849))/(18))))^(1/3) and now we have to go all the way back... not very friendly... as I mentioned before, it′s almost (or maybe absolutely) impossible to find a nicer exact form of w, so let me use the approximation w≈−.246266 v=((25)/4)+w≈6.00373 u=((√(25+4w))/2)≈2.45025 c=−(5/2)+((√(25+4w))/2)≈−.0497482 d=(√(((25)/2)−w−((131)/(2(√(25+4w))))))≈.787215i a=−(5/2)−((√(25+4w))/2)≈−4.95025 b=(√(−((25)/2)+w+((131)/(2(√(25+4w))))))≈5.11001i x_1 =a+bi≈−10.0603 x_2 =a−bi≈.159762 x_3 =c+d≈−.0497482+.787215i x_4 =c−d≈−.0497482−.787215i
$${x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{6}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{those}\:\mathrm{who}\:\mathrm{want}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}… \\ $$$$ \\ $$$$\left({x}−{a}−{b}\mathrm{i}\right)\left({x}−{a}+{b}\mathrm{i}\right)\left({x}−{c}−{d}\right)\left({x}−{c}+{d}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}\left({a}+{c}\right){x}^{\mathrm{3}} +\left({a}^{\mathrm{2}} +\mathrm{4}{ac}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right){x}^{\mathrm{2}} − \\ $$$$\:\:\:\:\:−\mathrm{2}\left({a}\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right){x}+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{1}.\:−\mathrm{2}\left({a}+{c}\right)=\mathrm{10} \\ $$$$\mathrm{2}.\:{a}^{\mathrm{2}} +\mathrm{4}{ac}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{d}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}.\:−\mathrm{2}\left({a}\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right)=\mathrm{6} \\ $$$$\mathrm{4}.\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)=−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{1}.\:{a}=−{c}−\mathrm{5}{e} \\ $$$$\mathrm{2}.\:{b}=\sqrt{−{a}^{\mathrm{2}} +\mathrm{4}{ac}−{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }=\sqrt{\mathrm{2}{c}^{\mathrm{2}} +\mathrm{10}{c}+{d}^{\mathrm{2}} −\mathrm{25}} \\ $$$$\mathrm{3}.\:{d}=\sqrt{{ac}+{c}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} {c}+\mathrm{3}}{{a}}}=\sqrt{−\frac{\mathrm{2}{c}^{\mathrm{3}} +\mathrm{15}{c}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}{c}+\mathrm{5}}} \\ $$$$\mathrm{2}.\:{b}=\sqrt{\frac{\mathrm{2}{c}^{\mathrm{3}} +\mathrm{15}{c}^{\mathrm{2}} −\mathrm{128}}{\mathrm{2}{c}+\mathrm{5}}} \\ $$$$\mathrm{4}. \\ $$$$\mathrm{4}{c}^{\mathrm{4}} +\mathrm{40}{c}^{\mathrm{3}} +\mathrm{75}{c}^{\mathrm{2}} −\mathrm{125}{c}+\frac{\mathrm{689}}{\mathrm{4}}−\frac{\mathrm{17161}}{\mathrm{4}\left(\mathrm{2}{c}+\mathrm{5}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{64}\left({c}^{\mathrm{6}} +\mathrm{15}{c}^{\mathrm{5}} +\mathrm{75}{c}^{\mathrm{4}} +\mathrm{125}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{2}} +\mathrm{20}{c}+\mathrm{1}\right)=\mathrm{0} \\ $$$${c}^{\mathrm{6}} +\mathrm{15}{c}^{\mathrm{5}} +\mathrm{75}{c}^{\mathrm{4}} +\mathrm{125}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{2}} +\mathrm{20}{c}+\mathrm{1}=\mathrm{0} \\ $$$${c}={u}−\frac{\mathrm{15}}{\mathrm{6}} \\ $$$${u}^{\mathrm{6}} −\frac{\mathrm{75}}{\mathrm{4}}{u}^{\mathrm{4}} +\frac{\mathrm{1939}}{\mathrm{16}}{u}^{\mathrm{2}} −\frac{\mathrm{17161}}{\mathrm{64}}=\mathrm{0} \\ $$$${u}=\sqrt{{v}} \\ $$$${v}^{\mathrm{3}} −\frac{\mathrm{75}}{\mathrm{4}}{v}^{\mathrm{2}} +\frac{\mathrm{1939}}{\mathrm{16}}{v}−\frac{\mathrm{17161}}{\mathrm{64}}=\mathrm{0} \\ $$$${v}={w}+\frac{\mathrm{75}}{\mathrm{12}} \\ $$$${w}^{\mathrm{3}} +\mathrm{4}{w}+\mathrm{1}=\mathrm{0} \\ $$$${w}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}+\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}= \\ $$$$=\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{849}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{849}}}{\mathrm{18}}} \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{go}\:\mathrm{all}\:\mathrm{the}\:\mathrm{way}\:\mathrm{back}… \\ $$$$\mathrm{not}\:\mathrm{very}\:\mathrm{friendly}… \\ $$$$\mathrm{as}\:\mathrm{I}\:\mathrm{mentioned}\:\mathrm{before},\:\mathrm{it}'\mathrm{s}\:\mathrm{almost}\:\left(\mathrm{or}\:\mathrm{maybe}\right. \\ $$$$\left.\mathrm{absolutely}\right)\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{nicer}\:\mathrm{exact} \\ $$$$\mathrm{form}\:\mathrm{of}\:{w},\:\mathrm{so}\:\mathrm{let}\:\mathrm{me}\:\mathrm{use}\:\mathrm{the}\:\mathrm{approximation} \\ $$$$ \\ $$$${w}\approx−.\mathrm{246266} \\ $$$${v}=\frac{\mathrm{25}}{\mathrm{4}}+{w}\approx\mathrm{6}.\mathrm{00373} \\ $$$${u}=\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx\mathrm{2}.\mathrm{45025} \\ $$$${c}=−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx−.\mathrm{0497482} \\ $$$${d}=\sqrt{\frac{\mathrm{25}}{\mathrm{2}}−{w}−\frac{\mathrm{131}}{\mathrm{2}\sqrt{\mathrm{25}+\mathrm{4}{w}}}}\approx.\mathrm{787215i} \\ $$$${a}=−\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx−\mathrm{4}.\mathrm{95025} \\ $$$${b}=\sqrt{−\frac{\mathrm{25}}{\mathrm{2}}+{w}+\frac{\mathrm{131}}{\mathrm{2}\sqrt{\mathrm{25}+\mathrm{4}{w}}}}\approx\mathrm{5}.\mathrm{11001i} \\ $$$${x}_{\mathrm{1}} ={a}+{b}\mathrm{i}\approx−\mathrm{10}.\mathrm{0603} \\ $$$${x}_{\mathrm{2}} ={a}−{b}\mathrm{i}\approx.\mathrm{159762} \\ $$$${x}_{\mathrm{3}} ={c}+{d}\approx−.\mathrm{0497482}+.\mathrm{787215i} \\ $$$${x}_{\mathrm{4}} ={c}−{d}\approx−.\mathrm{0497482}−.\mathrm{787215i} \\ $$
Commented by abdo mathsup 649 cc last updated on 30/May/18
thank you sir Mjs
$${thank}\:{you}\:{sir}\:{Mjs} \\ $$
Commented by Rasheed.Sindhi last updated on 30/May/18
^V Nice Sir!
$$\:^{\mathcal{V}} \mathcal{N}{ice}\:\mathcal{S}{ir}! \\ $$
Commented by behi83417@gmail.com last updated on 30/May/18
thanks in advance dear mr :MJS.
$${thanks}\:{in}\:{advance}\:{dear}\:{mr}\::{MJS}. \\ $$
Commented by MJS last updated on 30/May/18
you are welcome
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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