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Question Number 184787 by Mastermind last updated on 11/Jan/23

x^4 +16x^3 +9x^2 +256x+256=0 Find the values of x?

$$\mathrm{x}^{\mathrm{4}} +\mathrm{16x}^{\mathrm{3}} +\mathrm{9x}^{\mathrm{2}} +\mathrm{256x}+\mathrm{256}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}? \\ $$

Commented by MJS_new last updated on 11/Jan/23

I explained this many times before x^4 +ax^3 +bx^2 +cx+d=0 (1) try ±factors of the constant d if you don′t succeed: (2) let x=t−(a/4) to get t^4 +pt^2 +qt+r=0 we′re trying to find 2 square factors (t^2 −αt−β)(t^2 +αt−γ)=0 ⇔ t^4 −(α^2 +β+γ)t^2 +α(γ−β)t+βγ=0 by comparing the constants we get { ((−(α^2 +β+γ)=p)),((α(γ−β)=q)),((βγ=r)) :} solve (1) and (2) for β and γ then insert in (3) and transform to get (α^2 )^3 +j(α^2 )^2 +k(α^2 )+l=0 if this has got at least one nice solution for α^2 we get nice solutions for t^2 −αt−β=0 and t^2 +αt−γ=0 and also for x. if there′s no nice solution for α^2 it′s better to approximate for x from the given equation.

$$\mathrm{I}\:\mathrm{explained}\:\mathrm{this}\:\mathrm{many}\:\mathrm{times}\:\mathrm{before} \\ $$$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right) \\ $$$$\mathrm{try}\:\pm\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{constant}\:{d} \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{succeed}: \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{let}\:{x}={t}−\frac{{a}}{\mathrm{4}}\:\mathrm{to}\:\mathrm{get} \\ $$$${t}^{\mathrm{4}} +{pt}^{\mathrm{2}} +{qt}+{r}=\mathrm{0} \\ $$$$\mathrm{we}'\mathrm{re}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{find}\:\mathrm{2}\:\mathrm{square}\:\mathrm{factors} \\ $$$$\left({t}^{\mathrm{2}} −\alpha{t}−\beta\right)\left({t}^{\mathrm{2}} +\alpha{t}−\gamma\right)=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${t}^{\mathrm{4}} −\left(\alpha^{\mathrm{2}} +\beta+\gamma\right){t}^{\mathrm{2}} +\alpha\left(\gamma−\beta\right){t}+\beta\gamma=\mathrm{0} \\ $$$$\mathrm{by}\:\mathrm{comparing}\:\mathrm{the}\:\mathrm{constants}\:\mathrm{we}\:\mathrm{get} \\ $$$$\begin{cases}{−\left(\alpha^{\mathrm{2}} +\beta+\gamma\right)={p}}\\{\alpha\left(\gamma−\beta\right)={q}}\\{\beta\gamma={r}}\end{cases} \\ $$$$\mathrm{solve}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{for}\:\beta\:\mathrm{and}\:\gamma \\ $$$$\mathrm{then}\:\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{3}\right)\:\mathrm{and}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{get} \\ $$$$\left(\alpha^{\mathrm{2}} \right)^{\mathrm{3}} +{j}\left(\alpha^{\mathrm{2}} \right)^{\mathrm{2}} +{k}\left(\alpha^{\mathrm{2}} \right)+{l}=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{this}\:\mathrm{has}\:\mathrm{got}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{nice}\:\mathrm{solution}\:\mathrm{for} \\ $$$$\alpha^{\mathrm{2}} \:\mathrm{we}\:\mathrm{get}\:\mathrm{nice}\:\mathrm{solutions}\:\mathrm{for}\:{t}^{\mathrm{2}} −\alpha{t}−\beta=\mathrm{0}\:\mathrm{and} \\ $$$${t}^{\mathrm{2}} +\alpha{t}−\gamma=\mathrm{0}\:\mathrm{and}\:\mathrm{also}\:\mathrm{for}\:{x}.\:\mathrm{if}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{nice} \\ $$$$\mathrm{solution}\:\mathrm{for}\:\alpha^{\mathrm{2}} \:\mathrm{it}'\mathrm{s}\:\mathrm{better}\:\mathrm{to}\:\mathrm{approximate} \\ $$$$\mathrm{for}\:{x}\:\mathrm{from}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}. \\ $$

Commented by MJS_new last updated on 11/Jan/23

for the given equation we get x^4 +16x^3 +9x^2 +256x+256=0 x=t−4 t^4 +87t^2 +696t−1392=0 α=(√(87)) β=−4(√(87)) γ=4(√(87)) ... x_(1, 2) =−4−((√(87))/2)±((√(87+16(√(87))))/2) x_(3, 4) =−((8−(√(87)))/2)±((√(−87+16(√(87))))/2)i

$$\mathrm{for}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}^{\mathrm{4}} +\mathrm{16}{x}^{\mathrm{3}} +\mathrm{9}{x}^{\mathrm{2}} +\mathrm{256}{x}+\mathrm{256}=\mathrm{0} \\ $$$${x}={t}−\mathrm{4} \\ $$$${t}^{\mathrm{4}} +\mathrm{87}{t}^{\mathrm{2}} +\mathrm{696}{t}−\mathrm{1392}=\mathrm{0} \\ $$$$\alpha=\sqrt{\mathrm{87}} \\ $$$$\beta=−\mathrm{4}\sqrt{\mathrm{87}} \\ $$$$\gamma=\mathrm{4}\sqrt{\mathrm{87}} \\ $$$$... \\ $$$${x}_{\mathrm{1},\:\mathrm{2}} =−\mathrm{4}−\frac{\sqrt{\mathrm{87}}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{87}+\mathrm{16}\sqrt{\mathrm{87}}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =−\frac{\mathrm{8}−\sqrt{\mathrm{87}}}{\mathrm{2}}\pm\frac{\sqrt{−\mathrm{87}+\mathrm{16}\sqrt{\mathrm{87}}}}{\mathrm{2}}\mathrm{i} \\ $$

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