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x-4-ax-3-bx-2-cx-d-0-Find-x-




Question Number 105124 by ajfour last updated on 26/Jul/20
x^4 +ax^3 +bx^2 +cx+d=0  Find x.
$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${Find}\:{x}. \\ $$
Answered by ajfour last updated on 26/Jul/20
let  (x^2 +px+q)(x^2 +rx+m)=0  ⇒   p+r=a          q+m+pr=b          mp+qr = c             qm = d  Let   mp+(b−m−pr)r=c  ⇒     m = ((c−r(b−pr))/(p−r))  And    (b−pr−q)p+qr=c  ⇒      q=((c−p(b−pr))/(r−p))  ⇒   [((c−r(b−pr))/(p−r))][((c−p(b−pr))/(r−p))]=d  ⇒   c^2 −ac(b−pr)+pr(b−pr)^2                   +d(a^2 −4pr)=0  let   pr=t   ⇒  t^3 −2bt^2 +(b^2 +ac−4d)t+(c^2 −abc+a^2 d)      = 0  let   t=z+s  ⇒  (z^3 +3sz^2 +3s^2 z+s^3 )−2b(z^2 +2sz+s^2 )  +(b^2 +ac−4d)(z+s)+(c^2 −abc+a^2 d)=0  ⇒  z^3 +(3s−2b)z^2 +(3s^2 −4bs+b^2 +ac−4d)z  +[s^3 −2bs^2 +s(b^2 +ac−4d)+c^2 −abc+a^2 d]     = 0  If   3s=2b  ⇒  z^3 +(((4b^2 )/3)−((8b^2 )/3)+b^2 +ac−4d)z      +(((8b^3 )/(27))−((8b^3 )/9)+((2b^3 )/3)+((2abc)/3)−((8bd)/3)                   +c^2 −abc+a^2 d)=0  ⇒  z^3 +(ac−4d−(b^2 /3))z+(((2b^3 )/(27))−((abc)/3)−((8bd)/3)                                                   +c^2 +a^2 d) = 0  finding z by Cardano′s formula      t=pr= z+((2b)/3)  and  as  p+r=a  ⇒   p ,r = (a/2)±(√((a^2 /4)−z−((2b)/3)))         q= (d/m) = (d/([((c−r(b−pr))/(p−r))]))  Now,     x^2 +px+q = 0   ....
$${let}\:\:\left({x}^{\mathrm{2}} +{px}+{q}\right)\left({x}^{\mathrm{2}} +{rx}+{m}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{p}+{r}={a} \\ $$$$\:\:\:\:\:\:\:\:{q}+{m}+{pr}={b} \\ $$$$\:\:\:\:\:\:\:\:{mp}+{qr}\:=\:{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{qm}\:=\:{d} \\ $$$${Let}\:\:\:{mp}+\left({b}−{m}−{pr}\right){r}={c} \\ $$$$\Rightarrow\:\:\:\:\:{m}\:=\:\frac{{c}−{r}\left({b}−{pr}\right)}{{p}−{r}} \\ $$$${And}\:\:\:\:\left({b}−{pr}−{q}\right){p}+{qr}={c} \\ $$$$\Rightarrow\:\:\:\:\:\:{q}=\frac{{c}−{p}\left({b}−{pr}\right)}{{r}−{p}} \\ $$$$\Rightarrow\:\:\:\left[\frac{{c}−{r}\left({b}−{pr}\right)}{{p}−{r}}\right]\left[\frac{{c}−{p}\left({b}−{pr}\right)}{{r}−{p}}\right]={d} \\ $$$$\Rightarrow\:\:\:{c}^{\mathrm{2}} −{ac}\left({b}−{pr}\right)+{pr}\left({b}−{pr}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{d}\left({a}^{\mathrm{2}} −\mathrm{4}{pr}\right)=\mathrm{0} \\ $$$${let}\:\:\:{pr}={t}\:\:\:\Rightarrow \\ $$$${t}^{\mathrm{3}} −\mathrm{2}{bt}^{\mathrm{2}} +\left({b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right){t}+\left({c}^{\mathrm{2}} −{abc}+{a}^{\mathrm{2}} {d}\right) \\ $$$$\:\:\:\:=\:\mathrm{0} \\ $$$${let}\:\:\:{t}={z}+{s}\:\:\Rightarrow \\ $$$$\left({z}^{\mathrm{3}} +\mathrm{3}{sz}^{\mathrm{2}} +\mathrm{3}{s}^{\mathrm{2}} {z}+{s}^{\mathrm{3}} \right)−\mathrm{2}{b}\left({z}^{\mathrm{2}} +\mathrm{2}{sz}+{s}^{\mathrm{2}} \right) \\ $$$$+\left({b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right)\left({z}+{s}\right)+\left({c}^{\mathrm{2}} −{abc}+{a}^{\mathrm{2}} {d}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{3}{s}−\mathrm{2}{b}\right){z}^{\mathrm{2}} +\left(\mathrm{3}{s}^{\mathrm{2}} −\mathrm{4}{bs}+{b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right){z} \\ $$$$+\left[{s}^{\mathrm{3}} −\mathrm{2}{bs}^{\mathrm{2}} +{s}\left({b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right)+{c}^{\mathrm{2}} −{abc}+{a}^{\mathrm{2}} {d}\right] \\ $$$$\:\:\:=\:\mathrm{0} \\ $$$${If}\:\:\:\mathrm{3}{s}=\mathrm{2}{b}\:\:\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left(\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{8}{b}^{\mathrm{2}} }{\mathrm{3}}+{b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right){z} \\ $$$$\:\:\:\:+\left(\frac{\mathrm{8}{b}^{\mathrm{3}} }{\mathrm{27}}−\frac{\mathrm{8}{b}^{\mathrm{3}} }{\mathrm{9}}+\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{abc}}{\mathrm{3}}−\frac{\mathrm{8}{bd}}{\mathrm{3}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{c}^{\mathrm{2}} −{abc}+{a}^{\mathrm{2}} {d}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left({ac}−\mathrm{4}{d}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\right){z}+\left(\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}}−\frac{{abc}}{\mathrm{3}}−\frac{\mathrm{8}{bd}}{\mathrm{3}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{c}^{\mathrm{2}} +{a}^{\mathrm{2}} {d}\right)\:=\:\mathrm{0} \\ $$$${finding}\:{z}\:{by}\:{Cardano}'{s}\:{formula} \\ $$$$\:\:\:\:{t}={pr}=\:{z}+\frac{\mathrm{2}{b}}{\mathrm{3}} \\ $$$${and}\:\:{as}\:\:{p}+{r}={a} \\ $$$$\Rightarrow\:\:\:{p}\:,{r}\:=\:\frac{{a}}{\mathrm{2}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{z}−\frac{\mathrm{2}{b}}{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:{q}=\:\frac{{d}}{{m}}\:=\:\frac{{d}}{\left[\frac{{c}−{r}\left({b}−{pr}\right)}{{p}−{r}}\right]} \\ $$$${Now},\: \\ $$$$\:\:{x}^{\mathrm{2}} +{px}+{q}\:=\:\mathrm{0}\: \\ $$$$…. \\ $$

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