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Question Number 164019 by ajfour last updated on 13/Jan/22
x^4 +ax^3 +bx^2 +cx+d=0 ⇒ x^2 +(d/x^2 )+ax+(c/x)+b=0 say (1/x)=t ⇒ x^2 +ax+b+dt^2 +ct=0 let x^2 +ax+p=0 & dt^2 +ct+q=0 p+q=b x=−(a/2)±(√((a^2 /4)−p)) t=−(c/(2d))±(√((c^2 /(4d^2 ))−(q/d))) tx=1 ⇒ ((ac)/(4d))∓(a/2)(√((c^2 /(4d^2 ))−(q/d)))∓(c/(2d))(√((a^2 /4)−p)) +(√((a^2 /4)−p))(√((c^2 /(4d^2 ))−(q/d))) =1 let p=(a^2 /4) ⇒ ((ac)/(4d))∓(a/2)(√((c^2 /(4d^2 ))−(b/d)+(a^2 /(4d))))=1 ⇒ (c^2 /(4d^2 ))−(b/d)+(a^2 /(4d))=(4/a^2 )+(c^2 /(4d^2 ))−((2c)/(ad)) ⇒ 4a^2 b−a^4 +16d−8ac=0 if x=s+h s^4 +4hs(s^2 +h^2 )+6h^2 s^2 +h^4 +a{s^3 +3hs(s+h)+h^2 } +b{s^2 +2hs+h^2 }+c(s+h)+d=0 A=4h+a B=6h^2 +3ah+b C=4h^3 +3ah^2 +2bh+c D=h^4 +ah^3 +bh^2 +ch+d ⇒ (4h+a)^2 (8h^2 +4ah+4b−a^2 ) +16(h^4 +ah^3 +bh^2 +ch+d) =8(4h+a)(4h^3 +3ah^2 +2bh+c) .....
$$\:\:{x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\:\Rightarrow\:\:{x}^{\mathrm{2}} +\frac{{d}}{{x}^{\mathrm{2}} }+{ax}+\frac{{c}}{{x}}+{b}=\mathrm{0} \\ $$$${say}\:\:\:\frac{\mathrm{1}}{{x}}={t}\:\:\Rightarrow \\ $$$$\:\:{x}^{\mathrm{2}} +{ax}+{b}+{dt}^{\mathrm{2}} +{ct}=\mathrm{0} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}} +{ax}+{p}=\mathrm{0} \\ $$$$\&\:\:\:\:\:\:{dt}^{\mathrm{2}} +{ct}+{q}=\mathrm{0} \\ $$$$\:\:\:\:{p}+{q}={b} \\ $$$${x}=−\frac{{a}}{\mathrm{2}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}}\:\: \\ $$$${t}=−\frac{{c}}{\mathrm{2}{d}}\pm\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}} \\ $$$${tx}=\mathrm{1}\:\:\:\:\Rightarrow \\ $$$$\frac{{ac}}{\mathrm{4}{d}}\mp\frac{{a}}{\mathrm{2}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}}\mp\frac{{c}}{\mathrm{2}{d}}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}} \\ $$$$\:\:\:\:\:\:\:\:+\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}}\:=\mathrm{1} \\ $$$${let}\:\:\:{p}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\:\:\Rightarrow \\ $$$$\:\:\:\frac{{ac}}{\mathrm{4}{d}}\mp\frac{{a}}{\mathrm{2}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{b}}{{d}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{d}}}=\mathrm{1} \\ $$$$\Rightarrow\:\:\cancel{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }}−\frac{{b}}{{d}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{d}}=\frac{\mathrm{4}}{{a}^{\mathrm{2}} }+\cancel{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }}−\frac{\mathrm{2}{c}}{{ad}} \\ $$$$\Rightarrow\:\:\:\mathrm{4}{a}^{\mathrm{2}} {b}−{a}^{\mathrm{4}} +\mathrm{16}{d}−\mathrm{8}{ac}=\mathrm{0} \\ $$$${if}\:{x}={s}+{h} \\ $$$${s}^{\mathrm{4}} +\mathrm{4}{hs}\left({s}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)+\mathrm{6}{h}^{\mathrm{2}} {s}^{\mathrm{2}} +{h}^{\mathrm{4}} \\ $$$$+{a}\left\{{s}^{\mathrm{3}} +\mathrm{3}{hs}\left({s}+{h}\right)+{h}^{\mathrm{2}} \right\} \\ $$$$\:\:+{b}\left\{{s}^{\mathrm{2}} +\mathrm{2}{hs}+{h}^{\mathrm{2}} \right\}+{c}\left({s}+{h}\right)+{d}=\mathrm{0} \\ $$$${A}=\mathrm{4}{h}+{a} \\ $$$${B}=\mathrm{6}{h}^{\mathrm{2}} +\mathrm{3}{ah}+{b} \\ $$$${C}=\mathrm{4}{h}^{\mathrm{3}} +\mathrm{3}{ah}^{\mathrm{2}} +\mathrm{2}{bh}+{c} \\ $$$${D}={h}^{\mathrm{4}} +{ah}^{\mathrm{3}} +{bh}^{\mathrm{2}} +{ch}+{d} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{4}{h}+{a}\right)^{\mathrm{2}} \left(\mathrm{8}{h}^{\mathrm{2}} +\mathrm{4}{ah}+\mathrm{4}{b}−{a}^{\mathrm{2}} \right) \\ $$$$+\mathrm{16}\left({h}^{\mathrm{4}} +{ah}^{\mathrm{3}} +{bh}^{\mathrm{2}} +{ch}+{d}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{8}\left(\mathrm{4}{h}+{a}\right)\left(\mathrm{4}{h}^{\mathrm{3}} +\mathrm{3}{ah}^{\mathrm{2}} +\mathrm{2}{bh}+{c}\right) \\ $$$$….. \\ $$$$\:\:\:\:\:\: \\ $$

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