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 65271 by ajfour last updated on 27/Jul/19

x^4 +ax^2 +bx+c=0 solve for x.

$${x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${solve}\:{for}\:{x}. \\ $$

Answered by ajfour last updated on 29/Jul/19

let x^2 =px+t (px+t)^2 +a(px+t)+bx+c=0 p^2 (px+t)+t^2 +2ptx+a(px+t)+bx+c=0 ⇒ x(p^3 +2pt+ap+b) +t^2 +(p^2 +a)t+c=0 let coeff. of x be zero ⇒ t=−(((p^3 +ap+b))/(2p)) also then t^2 +(p^2 +a)t+c=0 ⇒ (p^3 +ap+b)^2 −2p(p^2 +a)(p^3 +ap+b)+4cp^2 =0 ⇒ p^6 +2ap^4 _(−) +2bp^3 _(−) +a^2 p^2 _(−) +2abp_(−) +b^2 −2p^6 −2ap^4 _(−) −2bp^3 _(−) −2ap^4 −2_− a^2 p^2 −2abp_(−) +4cp^2 =0 __________________________ p^6 +2ap^4 +(a^2 −4c)p^2 −b^2 =0 (its a cubic in p^2 ) Now t=−(((p^3 +ap+b))/(2p)) x^2 −px−t=0 _________________________■

$${let}\:{x}^{\mathrm{2}} ={px}+{t} \\ $$$$\left({px}+{t}\right)^{\mathrm{2}} +{a}\left({px}+{t}\right)+{bx}+{c}=\mathrm{0} \\ $$$${p}^{\mathrm{2}} \left({px}+{t}\right)+{t}^{\mathrm{2}} +\mathrm{2}{ptx}+{a}\left({px}+{t}\right)+{bx}+{c}=\mathrm{0} \\ $$$$\Rightarrow\:{x}\left({p}^{\mathrm{3}} +\mathrm{2}{pt}+{ap}+{b}\right) \\ $$$$\:\:\:\:\:\:+{t}^{\mathrm{2}} +\left({p}^{\mathrm{2}} +{a}\right){t}+{c}=\mathrm{0}\:\:\: \\ $$$${let}\:{coeff}.\:{of}\:{x}\:{be}\:{zero} \\ $$$$\Rightarrow\:\:\:{t}=−\frac{\left({p}^{\mathrm{3}} +{ap}+{b}\right)}{\mathrm{2}{p}} \\ $$$${also}\:\:{then} \\ $$$$\:\:\:\:\:\:\:\:{t}^{\mathrm{2}} +\left({p}^{\mathrm{2}} +{a}\right){t}+{c}=\mathrm{0} \\ $$$$\:\Rightarrow\:\:\left({p}^{\mathrm{3}} +{ap}+{b}\right)^{\mathrm{2}} \\ $$$$\:−\mathrm{2}{p}\left({p}^{\mathrm{2}} +{a}\right)\left({p}^{\mathrm{3}} +{ap}+{b}\right)+\mathrm{4}{cp}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{6}} +\underset{−} {\mathrm{2}{ap}^{\mathrm{4}} }+\underset{−} {\mathrm{2}{bp}^{\mathrm{3}} }+\underset{−} {{a}^{\mathrm{2}} {p}^{\mathrm{2}} }+\underset{−} {\mathrm{2}{abp}}+{b}^{\mathrm{2}} \\ $$$$−\mathrm{2}{p}^{\mathrm{6}} −\underset{−} {\mathrm{2}{ap}^{\mathrm{4}} }−\underset{−} {\mathrm{2}{bp}^{\mathrm{3}} }−\mathrm{2}{ap}^{\mathrm{4}} −\underset{−} {\mathrm{2}}{a}^{\mathrm{2}} {p}^{\mathrm{2}} \\ $$$$−\underset{−} {\mathrm{2}{abp}}+\mathrm{4}{cp}^{\mathrm{2}} =\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${p}^{\mathrm{6}} +\mathrm{2}{ap}^{\mathrm{4}} +\left({a}^{\mathrm{2}} −\mathrm{4}{c}\right){p}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({its}\:{a}\:{cubic}\:{in}\:{p}^{\mathrm{2}} \right) \\ $$$${Now}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{t}=−\frac{\left({p}^{\mathrm{3}} +{ap}+{b}\right)}{\mathrm{2}{p}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −{px}−{t}=\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\blacksquare \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com