Question Number 105124 by ajfour last updated on 26/Jul/20
$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${Find}\:{x}. \\ $$
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 ....](Q105149.png)
$${let}\:\:\left({x}^{\mathrm{2}} +{px}+{q}\right)\left({x}^{\mathrm{2}} +{rx}+{m}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{p}+{r}={a} \\ $$$$\:\:\:\:\:\:\:\:{q}+{m}+{pr}={b} \\ $$$$\:\:\:\:\:\:\:\:{mp}+{qr}\:=\:{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{qm}\:=\:{d} \\ $$$${Let}\:\:\:{mp}+\left({b}−{m}−{pr}\right){r}={c} \\ $$$$\Rightarrow\:\:\:\:\:{m}\:=\:\frac{{c}−{r}\left({b}−{pr}\right)}{{p}−{r}} \\ $$$${And}\:\:\:\:\left({b}−{pr}−{q}\right){p}+{qr}={c} \\ $$$$\Rightarrow\:\:\:\:\:\:{q}=\frac{{c}−{p}\left({b}−{pr}\right)}{{r}−{p}} \\ $$$$\Rightarrow\:\:\:\left[\frac{{c}−{r}\left({b}−{pr}\right)}{{p}−{r}}\right]\left[\frac{{c}−{p}\left({b}−{pr}\right)}{{r}−{p}}\right]={d} \\ $$$$\Rightarrow\:\:\:{c}^{\mathrm{2}} −{ac}\left({b}−{pr}\right)+{pr}\left({b}−{pr}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{d}\left({a}^{\mathrm{2}} −\mathrm{4}{pr}\right)=\mathrm{0} \\ $$$${let}\:\:\:{pr}={t}\:\:\:\Rightarrow \\ $$$${t}^{\mathrm{3}} −\mathrm{2}{bt}^{\mathrm{2}} +\left({b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right){t}+\left({c}^{\mathrm{2}} −{abc}+{a}^{\mathrm{2}} {d}\right) \\ $$$$\:\:\:\:=\:\mathrm{0} \\ $$$${let}\:\:\:{t}={z}+{s}\:\:\Rightarrow \\ $$$$\left({z}^{\mathrm{3}} +\mathrm{3}{sz}^{\mathrm{2}} +\mathrm{3}{s}^{\mathrm{2}} {z}+{s}^{\mathrm{3}} \right)−\mathrm{2}{b}\left({z}^{\mathrm{2}} +\mathrm{2}{sz}+{s}^{\mathrm{2}} \right) \\ $$$$+\left({b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right)\left({z}+{s}\right)+\left({c}^{\mathrm{2}} −{abc}+{a}^{\mathrm{2}} {d}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{3}{s}−\mathrm{2}{b}\right){z}^{\mathrm{2}} +\left(\mathrm{3}{s}^{\mathrm{2}} −\mathrm{4}{bs}+{b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right){z} \\ $$$$+\left[{s}^{\mathrm{3}} −\mathrm{2}{bs}^{\mathrm{2}} +{s}\left({b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right)+{c}^{\mathrm{2}} −{abc}+{a}^{\mathrm{2}} {d}\right] \\ $$$$\:\:\:=\:\mathrm{0} \\ $$$${If}\:\:\:\mathrm{3}{s}=\mathrm{2}{b}\:\:\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left(\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{8}{b}^{\mathrm{2}} }{\mathrm{3}}+{b}^{\mathrm{2}} +{ac}−\mathrm{4}{d}\right){z} \\ $$$$\:\:\:\:+\left(\frac{\mathrm{8}{b}^{\mathrm{3}} }{\mathrm{27}}−\frac{\mathrm{8}{b}^{\mathrm{3}} }{\mathrm{9}}+\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{abc}}{\mathrm{3}}−\frac{\mathrm{8}{bd}}{\mathrm{3}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{c}^{\mathrm{2}} −{abc}+{a}^{\mathrm{2}} {d}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left({ac}−\mathrm{4}{d}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\right){z}+\left(\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}}−\frac{{abc}}{\mathrm{3}}−\frac{\mathrm{8}{bd}}{\mathrm{3}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{c}^{\mathrm{2}} +{a}^{\mathrm{2}} {d}\right)\:=\:\mathrm{0} \\ $$$${finding}\:{z}\:{by}\:{Cardano}'{s}\:{formula} \\ $$$$\:\:\:\:{t}={pr}=\:{z}+\frac{\mathrm{2}{b}}{\mathrm{3}} \\ $$$${and}\:\:{as}\:\:{p}+{r}={a} \\ $$$$\Rightarrow\:\:\:{p}\:,{r}\:=\:\frac{{a}}{\mathrm{2}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{z}−\frac{\mathrm{2}{b}}{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:{q}=\:\frac{{d}}{{m}}\:=\:\frac{{d}}{\left[\frac{{c}−{r}\left({b}−{pr}\right)}{{p}−{r}}\right]} \\ $$$${Now},\: \\ $$$$\:\:{x}^{\mathrm{2}} +{px}+{q}\:=\:\mathrm{0}\: \\ $$$$.... \\ $$