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Question Number 139818 by Study last updated on 01/May/21

x^4 −22x^2 +x−114=0 x=?

$${x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{2}} +{x}−\mathrm{114}=\mathrm{0} \\ $$$${x}=? \\ $$

Answered by MJS_new last updated on 01/May/21

exact solution too complicated, approximation gives x_1 ≈−5.14749386 x_2 ≈5.11487664 x_(3, 4) ≈.0163086079±2.08076764i

$$\mathrm{exact}\:\mathrm{solution}\:\mathrm{too}\:\mathrm{complicated},\:\mathrm{approximation} \\ $$$$\mathrm{gives} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{5}.\mathrm{14749386} \\ $$$${x}_{\mathrm{2}} \approx\mathrm{5}.\mathrm{11487664} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} \approx.\mathrm{0163086079}\pm\mathrm{2}.\mathrm{08076764i} \\ $$

Commented by mr W last updated on 01/May/21

i guess the guy wanted to ask x^4 −22x^2 +x+114=0 but made again a typo and wrote x^4 −22x^2 +x−114=0.

$${i}\:{guess}\:{the}\:{guy}\:{wanted}\:{to}\:{ask} \\ $$$${x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{2}} +{x}+\mathrm{114}=\mathrm{0} \\ $$$${but}\:{made}\:{again}\:{a}\:{typo}\:{and}\:{wrote} \\ $$$${x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{2}} +{x}−\mathrm{114}=\mathrm{0}. \\ $$

Commented by MJS_new last updated on 01/May/21

I gave up guessing some time ago...

$$\mathrm{I}\:\mathrm{gave}\:\mathrm{up}\:\mathrm{guessing}\:\mathrm{some}\:\mathrm{time}\:\mathrm{ago}... \\ $$

Answered by MJS_new last updated on 01/May/21

generally x^4 +px^2 +qx+r=0 we try to find square factors (x^2 −αx−β)(x^2 +αx−γ)=0 by matching the constant factors. this leads to { ((−(α^2 +β+γ)=p)),((α(γ−β)=q)),((βγ=r)) :} solve the 1^(st) for β and the 2^(nd) for γ ⇒ β=−((α^3 +pα+q)/(2α))∧γ=−((α^3 +pα−q)/(2α)) inserting in the 3^(rd) gives ((α^6 +2pα^4 +p^2 α^2 −q^2 )/(4α^2 ))=r α^6 +2pα^4 +(p^2 −4r)α^2 −q^2 =0 let α=(√u) u^3 +2pu^2 +(p^2 −4r)u−q^2 =0 let u=v−((2p)/3) v^3 −((p^2 +12r)/3)v−((2p^3 −72pr+27q^2 )/(27))=0 if this has at least one solution v_1 ∈Q it makes insert backwards and exactly solve the given equation. otherwise it′s better to approximate

$$\mathrm{generally} \\ $$$${x}^{\mathrm{4}} +{px}^{\mathrm{2}} +{qx}+{r}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{square}\:\mathrm{factors} \\ $$$$\left({x}^{\mathrm{2}} −\alpha{x}−\beta\right)\left({x}^{\mathrm{2}} +\alpha{x}−\gamma\right)=\mathrm{0} \\ $$$$\mathrm{by}\:\mathrm{matching}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{factors}.\:\mathrm{this}\:\mathrm{leads} \\ $$$$\mathrm{to} \\ $$$$\begin{cases}{−\left(\alpha^{\mathrm{2}} +\beta+\gamma\right)={p}}\\{\alpha\left(\gamma−\beta\right)={q}}\\{\beta\gamma={r}}\end{cases} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{for}\:\beta\:\mathrm{and}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{for}\:\gamma\:\Rightarrow \\ $$$$\beta=−\frac{\alpha^{\mathrm{3}} +{p}\alpha+{q}}{\mathrm{2}\alpha}\wedge\gamma=−\frac{\alpha^{\mathrm{3}} +{p}\alpha−{q}}{\mathrm{2}\alpha} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{gives} \\ $$$$\frac{\alpha^{\mathrm{6}} +\mathrm{2}{p}\alpha^{\mathrm{4}} +{p}^{\mathrm{2}} \alpha^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{4}\alpha^{\mathrm{2}} }={r} \\ $$$$\alpha^{\mathrm{6}} +\mathrm{2}{p}\alpha^{\mathrm{4}} +\left({p}^{\mathrm{2}} −\mathrm{4}{r}\right)\alpha^{\mathrm{2}} −{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:\alpha=\sqrt{{u}} \\ $$$${u}^{\mathrm{3}} +\mathrm{2}{pu}^{\mathrm{2}} +\left({p}^{\mathrm{2}} −\mathrm{4}{r}\right){u}−{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:{u}={v}−\frac{\mathrm{2}{p}}{\mathrm{3}} \\ $$$${v}^{\mathrm{3}} −\frac{{p}^{\mathrm{2}} +\mathrm{12}{r}}{\mathrm{3}}{v}−\frac{\mathrm{2}{p}^{\mathrm{3}} −\mathrm{72}{pr}+\mathrm{27}{q}^{\mathrm{2}} }{\mathrm{27}}=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{this}\:\mathrm{has}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{solution}\:{v}_{\mathrm{1}} \in\mathbb{Q}\:\mathrm{it}\:\mathrm{makes} \\ $$$$\mathrm{insert}\:\mathrm{backwards}\:\mathrm{and}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equation}.\:\mathrm{otherwise}\:\mathrm{it}'\mathrm{s}\:\mathrm{better}\:\mathrm{to}\:\mathrm{approximate} \\ $$

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