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Question Number 65271 by ajfour last updated on 27/Jul/19

$$x^4+{ax}^2+bx+c=0$$$$solveforx.$$

Answered by ajfour last updated on 29/Jul/19

$$let{x}^2=px+t$$$${(px+t)}^2+a(px+t)+bx+c=0$$$$p^2(px+t)+t^2+2ptx+a(px+t)+bx+c=0$$$$\Rightarrow x(p^3+2pt+ap+b)$$$$+t^2+(p^2+a)t+c=0$$$$letcoeff.ofxbezero$$$$\Rightarrow t=-\frac{(p^3+ap+b)}{2p}$$$$alsothen$$$$t^2+(p^2+a)t+c=0$$$$\Rightarrow {(p^3+ap+b)}^2$$$$-2p(p^2+a)(p^3+ap+b)+4{cp}^2=0$$$$\Rightarrow$$$$p^6+\underset{-}{2{ap}^4}+\underset{-}{2{bp}^3}+\underset{-}{a^2{p}^2}+\underset{-}{2abp}+b^2$$$$-2p^6-\underset{-}{2{ap}^4}-\underset{-}{2{bp}^3}-2{ap}^4-\underset{-}{2}{a}^2{p}^2$$$$-\underset{-}{2abp}+4{cp}^2=0$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$p^6+2{ap}^4+(a^2-4c)p^2-b^2=0$$$$\left(itsacubicin{p}^2\right)$$$$Now$$$$t=-\frac{(p^3+ap+b)}{2p}$$$$x^2-px-t=0$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}◼$$

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