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Question-65830

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Question Number 65830 by ajfour last updated on 04/Aug/19
Commented by ajfour last updated on 04/Aug/19
x^4 +ax^3 +bx^2 +cx+d=0 let roots be −p,−q,−r,−s. Find p,q,r,s in terms of a,b,c,d. a,b,c,d ∈R.
$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${let}\:{roots}\:{be}\:−{p},−{q},−{r},−{s}. \\ $$$${Find}\:{p},{q},{r},{s}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d}. \\ $$$${a},{b},{c},{d}\:\in\mathbb{R}. \\ $$
Commented by MJS last updated on 05/Aug/19
my method as I did before is putting −p=α−(√β) −q=α+(√β) −r=γ−(√δ) −s=γ+(√δ) which leads to α=−γ−(a/2) β=−γ^2 −(a/2)γ+((a^2 −4b)/8)+((a^3 −4ab+8c)/(8(4γ+a))) δ=−γ^2 −(a/2)γ+((a^2 −4b)/8)−((a^3 −4ab+8c)/(8(4γ+a))) γ^6 +((3a)/2)γ^5 +((3a^2 +2b)/4)γ^4 +((a(a^2 +4b))/8)γ^3 +((2a^2 b+ac+b^2 −4d)/(16))γ^2 +((a(ac+b^2 −4d))/(32))γ−((a^2 d−abc+c^2 )/(64))=0 γ=r−(a/4) r^6 −((3a^2 −8b)/(16))r^4 +((3a^4 −16a^2 b+16ac+16(b^2 −4d))/(256))r^2 −(((a^3 −4ab+8c)^2 )/(4096))=0 r=(√s) s^3 −((3a^2 −8b)/(16))s^2 +((3a^4 −16a^2 b+16ac+16(b^2 −4d))/(256))s−(((a^3 −4ab+8c)^2 )/(4096))=0 s=t+((3a^2 −8b)/(48)) t^3 +((3ac−b^2 −12d)/(48))t−((27a^2 d−9abc+2b^3 −72bd+27c^2 )/(1728))=0 and we can always solve this depending on D we need Cardano′s or the trigonometric method D=27a^4 d^2 −18a^3 bcd+4a^3 c^3 +4a^2 b^3 d−a^2 b^2 c^2 −144a^2 bd^2 +6a^2 c^2 d+80ab^2 cd−18abc^3 +192acd^2 −16b^4 d+4b^3 c^2 +128b^2 d^2 −144bc^2 d+27c^4 −256d^3
$$\mathrm{my}\:\mathrm{method}\:\mathrm{as}\:\mathrm{I}\:\mathrm{did}\:\mathrm{before}\:\mathrm{is}\:\mathrm{putting} \\ $$$$−{p}=\alpha−\sqrt{\beta} \\ $$$$−{q}=\alpha+\sqrt{\beta} \\ $$$$−{r}=\gamma−\sqrt{\delta} \\ $$$$−{s}=\gamma+\sqrt{\delta} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\alpha=−\gamma−\frac{{a}}{\mathrm{2}} \\ $$$$\beta=−\gamma^{\mathrm{2}} −\frac{{a}}{\mathrm{2}}\gamma+\frac{{a}^{\mathrm{2}} −\mathrm{4}{b}}{\mathrm{8}}+\frac{{a}^{\mathrm{3}} −\mathrm{4}{ab}+\mathrm{8}{c}}{\mathrm{8}\left(\mathrm{4}\gamma+{a}\right)} \\ $$$$\delta=−\gamma^{\mathrm{2}} −\frac{{a}}{\mathrm{2}}\gamma+\frac{{a}^{\mathrm{2}} −\mathrm{4}{b}}{\mathrm{8}}−\frac{{a}^{\mathrm{3}} −\mathrm{4}{ab}+\mathrm{8}{c}}{\mathrm{8}\left(\mathrm{4}\gamma+{a}\right)} \\ $$$$\gamma^{\mathrm{6}} +\frac{\mathrm{3}{a}}{\mathrm{2}}\gamma^{\mathrm{5}} +\frac{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{b}}{\mathrm{4}}\gamma^{\mathrm{4}} +\frac{{a}\left({a}^{\mathrm{2}} +\mathrm{4}{b}\right)}{\mathrm{8}}\gamma^{\mathrm{3}} +\frac{\mathrm{2}{a}^{\mathrm{2}} {b}+{ac}+{b}^{\mathrm{2}} −\mathrm{4}{d}}{\mathrm{16}}\gamma^{\mathrm{2}} +\frac{{a}\left({ac}+{b}^{\mathrm{2}} −\mathrm{4}{d}\right)}{\mathrm{32}}\gamma−\frac{{a}^{\mathrm{2}} {d}−{abc}+{c}^{\mathrm{2}} }{\mathrm{64}}=\mathrm{0} \\ $$$$\gamma={r}−\frac{{a}}{\mathrm{4}} \\ $$$${r}^{\mathrm{6}} −\frac{\mathrm{3}{a}^{\mathrm{2}} −\mathrm{8}{b}}{\mathrm{16}}{r}^{\mathrm{4}} +\frac{\mathrm{3}{a}^{\mathrm{4}} −\mathrm{16}{a}^{\mathrm{2}} {b}+\mathrm{16}{ac}+\mathrm{16}\left({b}^{\mathrm{2}} −\mathrm{4}{d}\right)}{\mathrm{256}}{r}^{\mathrm{2}} −\frac{\left({a}^{\mathrm{3}} −\mathrm{4}{ab}+\mathrm{8}{c}\right)^{\mathrm{2}} }{\mathrm{4096}}=\mathrm{0} \\ $$$${r}=\sqrt{{s}} \\ $$$${s}^{\mathrm{3}} −\frac{\mathrm{3}{a}^{\mathrm{2}} −\mathrm{8}{b}}{\mathrm{16}}{s}^{\mathrm{2}} +\frac{\mathrm{3}{a}^{\mathrm{4}} −\mathrm{16}{a}^{\mathrm{2}} {b}+\mathrm{16}{ac}+\mathrm{16}\left({b}^{\mathrm{2}} −\mathrm{4}{d}\right)}{\mathrm{256}}{s}−\frac{\left({a}^{\mathrm{3}} −\mathrm{4}{ab}+\mathrm{8}{c}\right)^{\mathrm{2}} }{\mathrm{4096}}=\mathrm{0} \\ $$$${s}={t}+\frac{\mathrm{3}{a}^{\mathrm{2}} −\mathrm{8}{b}}{\mathrm{48}} \\ $$$${t}^{\mathrm{3}} +\frac{\mathrm{3}{ac}−{b}^{\mathrm{2}} −\mathrm{12}{d}}{\mathrm{48}}{t}−\frac{\mathrm{27}{a}^{\mathrm{2}} {d}−\mathrm{9}{abc}+\mathrm{2}{b}^{\mathrm{3}} −\mathrm{72}{bd}+\mathrm{27}{c}^{\mathrm{2}} }{\mathrm{1728}}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{always}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\mathrm{depending}\:\mathrm{on}\:{D}\:\mathrm{we}\:\mathrm{need}\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{or}\:\mathrm{the} \\ $$$$\mathrm{trigonometric}\:\mathrm{method} \\ $$$${D}=\mathrm{27}{a}^{\mathrm{4}} {d}^{\mathrm{2}} −\mathrm{18}{a}^{\mathrm{3}} {bcd}+\mathrm{4}{a}^{\mathrm{3}} {c}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{3}} {d}−{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{144}{a}^{\mathrm{2}} {bd}^{\mathrm{2}} +\mathrm{6}{a}^{\mathrm{2}} {c}^{\mathrm{2}} {d}+\mathrm{80}{ab}^{\mathrm{2}} {cd}−\mathrm{18}{abc}^{\mathrm{3}} +\mathrm{192}{acd}^{\mathrm{2}} −\mathrm{16}{b}^{\mathrm{4}} {d}+\mathrm{4}{b}^{\mathrm{3}} {c}^{\mathrm{2}} +\mathrm{128}{b}^{\mathrm{2}} {d}^{\mathrm{2}} −\mathrm{144}{bc}^{\mathrm{2}} {d}+\mathrm{27}{c}^{\mathrm{4}} −\mathrm{256}{d}^{\mathrm{3}} \\ $$
Commented by ajfour last updated on 05/Aug/19
my method is now simpler still; some calculation error lead me to disbelieve.
$${my}\:{method}\:{is}\:{now}\:{simpler}\:{still}; \\ $$$${some}\:{calculation}\:{error}\:{lead}\:{me} \\ $$$${to}\:{disbelieve}. \\ $$
Answered by ajfour last updated on 05/Aug/19
x^4 +ax^3 +bx^2 +cx+d=0 x=t−(a/4) t^4 +At^2 +Bt+C=0 t^2 −Pt+(1/2)(A+P^2 +(B/P))=0 ...(I) P^( 6) +2AP^( 4) +(A^2 −4C)P^( 2) −B^2 =0 Cardano′s or trigonometric sol^n provides for P^( 2) . D=(1/4)(((2A^3 )/(27))+B^2 −((8AC)/3))^2 −(1/(27))((A^2 /3)+4C)^3 Any value of P^( 2) provides all the four roots using eq.(I).
$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${x}={t}−\frac{{a}}{\mathrm{4}} \\ $$$${t}^{\mathrm{4}} +{At}^{\mathrm{2}} +{Bt}+{C}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −{Pt}+\frac{\mathrm{1}}{\mathrm{2}}\left({A}+{P}^{\mathrm{2}} +\frac{{B}}{{P}}\right)=\mathrm{0}\:\:\:\:…\left({I}\right) \\ $$$${P}^{\:\mathrm{6}} +\mathrm{2}{AP}^{\:\mathrm{4}} +\left({A}^{\mathrm{2}} −\mathrm{4}{C}\right){P}^{\:\mathrm{2}} −{B}^{\mathrm{2}} =\mathrm{0} \\ $$$${Cardano}'{s}\:{or}\:{trigonometric}\:{sol}^{{n}} \\ $$$${provides}\:{for}\:{P}^{\:\mathrm{2}} . \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{2}{A}^{\mathrm{3}} }{\mathrm{27}}+{B}^{\mathrm{2}} −\frac{\mathrm{8}{AC}}{\mathrm{3}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}\left(\frac{{A}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{4}{C}\right)^{\mathrm{3}} \\ $$$${Any}\:{value}\:{of}\:{P}^{\:\mathrm{2}} \:{provides}\:{all}\:{the} \\ $$$${four}\:{roots}\:{using}\:{eq}.\left({I}\right). \\ $$
Commented by ajfour last updated on 05/Aug/19
example: t^4 −15t^2 −10t+24=0 P^( 6) −30P^( 4) +129P^( 2) −100=0 P^2 =1,4,25 ⇒ t^2 ±t−7±5=0 or t^2 ±2t−((11)/2)±(5/2)=0 or t^2 ±5t+5±1=0 from each pairs roots are t=1,−2,−3,4 (quite true).
$${example}: \\ $$$${t}^{\mathrm{4}} −\mathrm{15}{t}^{\mathrm{2}} −\mathrm{10}{t}+\mathrm{24}=\mathrm{0} \\ $$$${P}^{\:\mathrm{6}} −\mathrm{30}{P}^{\:\mathrm{4}} +\mathrm{129}{P}^{\:\mathrm{2}} −\mathrm{100}=\mathrm{0} \\ $$$${P}\:^{\mathrm{2}} =\mathrm{1},\mathrm{4},\mathrm{25} \\ $$$$\Rightarrow \\ $$$${t}^{\mathrm{2}} \pm{t}−\mathrm{7}\pm\mathrm{5}=\mathrm{0}\:\:\:\:\:\:{or} \\ $$$${t}^{\mathrm{2}} \pm\mathrm{2}{t}−\frac{\mathrm{11}}{\mathrm{2}}\pm\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{0}\:\:{or} \\ $$$${t}^{\mathrm{2}} \pm\mathrm{5}{t}+\mathrm{5}\pm\mathrm{1}=\mathrm{0} \\ $$$${from}\:{each}\:{pairs}\:\:{roots}\:{are} \\ $$$${t}=\mathrm{1},−\mathrm{2},−\mathrm{3},\mathrm{4}\:\:\:\:\:\left({quite}\:{true}\right). \\ $$
Commented by MJS last updated on 05/Aug/19
yes. but also p=−p are there 2 equations of the 6 possible for t which generally give the 4 solutions? can you determine which pairs of equations will do?
$$\mathrm{yes}. \\ $$$$\mathrm{but}\:\mathrm{also}\:{p}=−{p} \\ $$$$\mathrm{are}\:\mathrm{there}\:\mathrm{2}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{6}\:\mathrm{possible}\:\mathrm{for}\:{t} \\ $$$$\mathrm{which}\:\mathrm{generally}\:\mathrm{give}\:\mathrm{the}\:\mathrm{4}\:\mathrm{solutions}? \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{determine}\:\mathrm{which}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\mathrm{will}\:\mathrm{do}? \\ $$
Commented by ajfour last updated on 05/Aug/19
[t^2 −Pt+(1/2)(A+P^( 2) +(B/P))][t^2 +Pt+(1/2)(A+P^( 2) −(B/P))]=0 we can choose just one value of P and use both equations, or P =±(√P^( 2) ) in any one of them, is that what you asked sir? its all the same, natural beauty!
$$\left[{t}^{\mathrm{2}} −{Pt}+\frac{\mathrm{1}}{\mathrm{2}}\left({A}+{P}^{\:\mathrm{2}} +\frac{{B}}{{P}}\right)\right]\left[{t}^{\mathrm{2}} +{Pt}+\frac{\mathrm{1}}{\mathrm{2}}\left({A}+{P}^{\:\mathrm{2}} −\frac{{B}}{{P}}\right)\right]=\mathrm{0} \\ $$$${we}\:{can}\:{choose}\:{just}\:{one}\:{value}\:{of} \\ $$$${P}\:{and}\:{use}\:{both}\:{equations},\:{or} \\ $$$${P}\:=\pm\sqrt{{P}^{\:\mathrm{2}} }\:\:{in}\:{any}\:{one}\:{of}\:{them}, \\ $$$${is}\:{that}\:{what}\:{you}\:{asked}\:{sir}? \\ $$$${its}\:{all}\:{the}\:{same},\:{natural}\:{beauty}! \\ $$

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