Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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)  .....

$$\:\:{x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$ $$\:\Rightarrow\:\:{x}^{\mathrm{2}} +\frac{{d}}{{x}^{\mathrm{2}} }+{ax}+\frac{{c}}{{x}}+{b}=\mathrm{0} \\ $$ $${say}\:\:\:\frac{\mathrm{1}}{{x}}={t}\:\:\Rightarrow \\ $$ $$\:\:{x}^{\mathrm{2}} +{ax}+{b}+{dt}^{\mathrm{2}} +{ct}=\mathrm{0} \\ $$ $${let}\:\:\:\:{x}^{\mathrm{2}} +{ax}+{p}=\mathrm{0} \\ $$ $$\&\:\:\:\:\:\:{dt}^{\mathrm{2}} +{ct}+{q}=\mathrm{0} \\ $$ $$\:\:\:\:{p}+{q}={b} \\ $$ $${x}=−\frac{{a}}{\mathrm{2}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}}\:\: \\ $$ $${t}=−\frac{{c}}{\mathrm{2}{d}}\pm\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}} \\ $$ $${tx}=\mathrm{1}\:\:\:\:\Rightarrow \\ $$ $$\frac{{ac}}{\mathrm{4}{d}}\mp\frac{{a}}{\mathrm{2}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}}\mp\frac{{c}}{\mathrm{2}{d}}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}} \\ $$ $$\:\:\:\:\:\:\:\:+\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}}\:=\mathrm{1} \\ $$ $${let}\:\:\:{p}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\:\:\Rightarrow \\ $$ $$\:\:\:\frac{{ac}}{\mathrm{4}{d}}\mp\frac{{a}}{\mathrm{2}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{b}}{{d}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{d}}}=\mathrm{1} \\ $$ $$\Rightarrow\:\:\cancel{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }}−\frac{{b}}{{d}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{d}}=\frac{\mathrm{4}}{{a}^{\mathrm{2}} }+\cancel{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }}−\frac{\mathrm{2}{c}}{{ad}} \\ $$ $$\Rightarrow\:\:\:\mathrm{4}{a}^{\mathrm{2}} {b}−{a}^{\mathrm{4}} +\mathrm{16}{d}−\mathrm{8}{ac}=\mathrm{0} \\ $$ $${if}\:{x}={s}+{h} \\ $$ $${s}^{\mathrm{4}} +\mathrm{4}{hs}\left({s}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)+\mathrm{6}{h}^{\mathrm{2}} {s}^{\mathrm{2}} +{h}^{\mathrm{4}} \\ $$ $$+{a}\left\{{s}^{\mathrm{3}} +\mathrm{3}{hs}\left({s}+{h}\right)+{h}^{\mathrm{2}} \right\} \\ $$ $$\:\:+{b}\left\{{s}^{\mathrm{2}} +\mathrm{2}{hs}+{h}^{\mathrm{2}} \right\}+{c}\left({s}+{h}\right)+{d}=\mathrm{0} \\ $$ $${A}=\mathrm{4}{h}+{a} \\ $$ $${B}=\mathrm{6}{h}^{\mathrm{2}} +\mathrm{3}{ah}+{b} \\ $$ $${C}=\mathrm{4}{h}^{\mathrm{3}} +\mathrm{3}{ah}^{\mathrm{2}} +\mathrm{2}{bh}+{c} \\ $$ $${D}={h}^{\mathrm{4}} +{ah}^{\mathrm{3}} +{bh}^{\mathrm{2}} +{ch}+{d} \\ $$ $$\Rightarrow \\ $$ $$\left(\mathrm{4}{h}+{a}\right)^{\mathrm{2}} \left(\mathrm{8}{h}^{\mathrm{2}} +\mathrm{4}{ah}+\mathrm{4}{b}−{a}^{\mathrm{2}} \right) \\ $$ $$+\mathrm{16}\left({h}^{\mathrm{4}} +{ah}^{\mathrm{3}} +{bh}^{\mathrm{2}} +{ch}+{d}\right) \\ $$ $$\:\:\:\:\:\:=\mathrm{8}\left(\mathrm{4}{h}+{a}\right)\left(\mathrm{4}{h}^{\mathrm{3}} +\mathrm{3}{ah}^{\mathrm{2}} +\mathrm{2}{bh}+{c}\right) \\ $$ $$..... \\ $$ $$\:\:\:\:\:\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com