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Question Number 36219 by MJS last updated on 30/May/18

in question 34992 this had to be solved: t^4 +8t^3 +2t^2 −8t+1=0 here another concept worked (I tried some) t_1 =a+b+c+d t_2 =a+b−c−d t_3 =a−b+c−d t_3 =a−b−c+d (t−t_1 )(t−t_2 )(t−t_3 )(t−t_4 )=0 leads to t^4 −4at^3 +2(3a^2 −b^2 −c^2 −d^2 )t^2 + +4(a(−a^2 +b^2 +c^2 +d^2 )−2bcd)t+ +a^4 +b^4 +c^4 +d^4 −2(a^2 (b^2 +c^2 +d^2 )+b^2 (c^2 +d^2 )+c^2 d^2 ) 1. −4a=8 2. 2(3a^2 −b^2 −c^2 −d^2 )=2 3. 4(a(−a^2 +b^2 +c^2 +d^2 )−2bcd)=−8 4. a^4 +b^4 +c^4 +d^4 −2(a^2 (b^2 +c^2 +d^2 )+b^2 (c^2 +d^2 )+c^2 d^2 )=1 1. a=−2 2. b^2 +c^2 +d^2 =11 3. b^2 +c^2 +d^2 +bcd=5 4. ... 3. −2. bcd=−6 now I tried by chance if there are easy real solutions b^2 =p, c^2 =q, d^2 =r, 0≤p≤q≤r p+q+r=11 this leads to p=2, q=3, p=6 ∣b∣=(√2), ∣c∣=(√3), ∣d∣=(√6) with one or three out of a, b, c negative because of bcd=−6 because of the nature of t_1 , t_2 , t_3 , t_4 these lead to the same solutions I took b=−(√2), c=−(√3), d=−(√6) t_1 =−2−(√2)−(√3)−(√6) t_2 =−2−(√2)+(√3)+(√6) t_3 =−2+(√2)−(√3)+(√6) t_4 =−2+(√2)+(√3)−(√6)

$$\mathrm{in}\:\mathrm{question}\:\mathrm{34992}\:\mathrm{this}\:\mathrm{had}\:\mathrm{to}\:\mathrm{be}\:\mathrm{solved}: \\ $$$${t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{here}\:\mathrm{another}\:\mathrm{concept}\:\mathrm{worked}\:\left(\mathrm{I}\:\mathrm{tried}\:\mathrm{some}\right) \\ $$$${t}_{\mathrm{1}} ={a}+{b}+{c}+{d} \\ $$$${t}_{\mathrm{2}} ={a}+{b}−{c}−{d} \\ $$$${t}_{\mathrm{3}} ={a}−{b}+{c}−{d} \\ $$$${t}_{\mathrm{3}} ={a}−{b}−{c}+{d} \\ $$$$\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)\left({t}−{t}_{\mathrm{3}} \right)\left({t}−{t}_{\mathrm{4}} \right)=\mathrm{0} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$${t}^{\mathrm{4}} −\mathrm{4}{at}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right){t}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:+\mathrm{4}\left({a}\left(−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)−\mathrm{2}{bcd}\right){t}+ \\ $$$$\:\:\:\:\:+{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} −\mathrm{2}\left({a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {d}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{1}.\:−\mathrm{4}{a}=\mathrm{8} \\ $$$$\mathrm{2}.\:\mathrm{2}\left(\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)=\mathrm{2} \\ $$$$\mathrm{3}.\:\mathrm{4}\left({a}\left(−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)−\mathrm{2}{bcd}\right)=−\mathrm{8} \\ $$$$\mathrm{4}.\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} −\mathrm{2}\left({a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {d}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{1}.\:{a}=−\mathrm{2} \\ $$$$\mathrm{2}.\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{11} \\ $$$$\mathrm{3}.\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{bcd}=\mathrm{5} \\ $$$$\mathrm{4}.\:... \\ $$$$\mathrm{3}.\:−\mathrm{2}.\:{bcd}=−\mathrm{6} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{I}\:\mathrm{tried}\:\mathrm{by}\:\mathrm{chance}\:\mathrm{if}\:\mathrm{there}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{real}\:\mathrm{solutions} \\ $$$${b}^{\mathrm{2}} ={p},\:{c}^{\mathrm{2}} ={q},\:{d}^{\mathrm{2}} ={r},\:\mathrm{0}\leqslant{p}\leqslant{q}\leqslant{r} \\ $$$${p}+{q}+{r}=\mathrm{11} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$${p}=\mathrm{2},\:{q}=\mathrm{3},\:{p}=\mathrm{6} \\ $$$$\mid{b}\mid=\sqrt{\mathrm{2}},\:\mid{c}\mid=\sqrt{\mathrm{3}},\:\mid{d}\mid=\sqrt{\mathrm{6}}\:\mathrm{with}\:\mathrm{one}\:\mathrm{or}\:\mathrm{three}\:\mathrm{out} \\ $$$$\mathrm{of}\:{a},\:{b},\:{c}\:\mathrm{negative}\:\mathrm{because}\:\mathrm{of}\:{bcd}=−\mathrm{6} \\ $$$$\mathrm{because}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:{t}_{\mathrm{1}} ,\:{t}_{\mathrm{2}} ,\:{t}_{\mathrm{3}} ,\:{t}_{\mathrm{4}} \:\mathrm{these} \\ $$$$\mathrm{lead}\:\mathrm{to}\:\mathrm{the}\:\mathrm{same}\:\mathrm{solutions} \\ $$$$\mathrm{I}\:\mathrm{took}\:{b}=−\sqrt{\mathrm{2}},\:{c}=−\sqrt{\mathrm{3}},\:{d}=−\sqrt{\mathrm{6}} \\ $$$${t}_{\mathrm{1}} =−\mathrm{2}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\sqrt{\mathrm{6}} \\ $$$${t}_{\mathrm{2}} =−\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$$${t}_{\mathrm{3}} =−\mathrm{2}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$$${t}_{\mathrm{4}} =−\mathrm{2}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{6}} \\ $$

Commented by Rasheed.Sindhi last updated on 30/May/18

Creative work , I think! �� �� ✌

Commented by tanmay.chaudhury50@gmail.com last updated on 31/May/18

t^4 +8t^3 +2t^2 −8t+1=0 t^2 +8t+2−(8/t)+(1/t^2 )=0 t^2 +(1/t^2 )+8(t−(1/t))+2=0 (t−(1/t))^2 +2+8(t−(1/t))+2=0 k=t−(1/t) k^2 +8k+4=0 k=((−8±(√(64−16)) )/2) =((−8±4(√3) )/2) =−4±2(√3) k=−4+2(√3) and −4−2(√3) now t−(1/t)=k t^2 −1=tk t^2 −tk−1=0 t=((k±(√(k^2 +4)) )/2) =((k+(√(k^2 +4)) )/2) and ((k−(√(k^2 +4)) )/2) now put value of k=−4+2(√3) t=((−4+2(√3) ±(√(16−16(√3) +12+4)))/2) t=((−4+2(√3) ±(√(32−16(√3) )))/2) t=−2+(√3) ±(√(8−4(√3) )) t=−2+(√3) +(√(8−4(√3) )) and−2+(√3) −(√(8−4(√3) )) now put k=−4−2(√3) you will get two value of t so total 2+2=4 roots of t found

$${t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} +\mathrm{8}{t}+\mathrm{2}−\frac{\mathrm{8}}{{t}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\mathrm{0} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{8}\left({t}−\frac{\mathrm{1}}{{t}}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}+\mathrm{8}\left({t}−\frac{\mathrm{1}}{{t}}\right)+\mathrm{2}=\mathrm{0} \\ $$$${k}={t}−\frac{\mathrm{1}}{{t}} \\ $$$${k}^{\mathrm{2}} +\mathrm{8}{k}+\mathrm{4}=\mathrm{0} \\ $$$${k}=\frac{−\mathrm{8}\pm\sqrt{\mathrm{64}−\mathrm{16}}\:}{\mathrm{2}} \\ $$$$=\frac{−\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{3}}\:}{\mathrm{2}} \\ $$$$=−\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$$${k}=−\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\:{and}\:−\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\:\: \\ $$$${now} \\ $$$${t}−\frac{\mathrm{1}}{{t}}={k} \\ $$$${t}^{\mathrm{2}} −\mathrm{1}={tk} \\ $$$${t}^{\mathrm{2}} −{tk}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{{k}\pm\sqrt{{k}^{\mathrm{2}} +\mathrm{4}}\:}{\mathrm{2}} \\ $$$$=\frac{{k}+\sqrt{{k}^{\mathrm{2}} +\mathrm{4}}\:}{\mathrm{2}}\:{and}\:\frac{{k}−\sqrt{{k}^{\mathrm{2}} +\mathrm{4}}\:}{\mathrm{2}} \\ $$$${now}\:{put}\:{value}\:{of}\:{k}=−\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\: \\ $$$${t}=\frac{−\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\pm\sqrt{\mathrm{16}−\mathrm{16}\sqrt{\mathrm{3}}\:+\mathrm{12}+\mathrm{4}}}{\mathrm{2}} \\ $$$${t}=\frac{−\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\pm\sqrt{\mathrm{32}−\mathrm{16}\sqrt{\mathrm{3}}\:\:}}{\mathrm{2}} \\ $$$${t}=−\mathrm{2}+\sqrt{\mathrm{3}}\:\pm\sqrt{\mathrm{8}−\mathrm{4}\sqrt{\mathrm{3}}\:} \\ $$$${t}=−\mathrm{2}+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{8}−\mathrm{4}\sqrt{\mathrm{3}}\:}\:\:{and}−\mathrm{2}+\sqrt{\mathrm{3}}\:\:−\sqrt{\mathrm{8}−\mathrm{4}\sqrt{\mathrm{3}}\:\:} \\ $$$${now}\:{put}\:{k}=−\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\:\:{you}\:{will}\:{get}\:{two}\:{value}\:{of} \\ $$$${t}\:{so}\:{total}\:\mathrm{2}+\mathrm{2}=\mathrm{4}\:{roots}\:{of}\:{t}\:{found} \\ $$

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