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Solve-in-R-x-2-4x-40x-2-32x-16-

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Question Number 156809 by MathSh last updated on 15/Oct/21
Solve in R x^2 + 4x = (√(40x^2 + 32x - 16))
$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:=\:\sqrt{\mathrm{40x}^{\mathrm{2}} \:+\:\mathrm{32x}\:-\:\mathrm{16}} \\ $$
Commented by MathSh last updated on 15/Oct/21
Yes dear Ser
$$\mathrm{Yes}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Answered by Rasheed.Sindhi last updated on 15/Oct/21
x(x+4)= (√(40x^2 + 32x + 16)) x^2 (x+4)^2 =40x^2 + 32x + 16 x^4 +8x^3 +16x^2 −40x^2 − 32x − 16=0 x^4 +8x^3 −24x^2 − 32x − 16=0 x=3.1850,−10.0814 (Using calculator) You can see exact results in an advanced calculator by yourself. They′re too complicated!
$$\:\mathrm{x}\left(\mathrm{x}+\mathrm{4}\right)=\:\sqrt{\mathrm{40x}^{\mathrm{2}} \:+\:\mathrm{32x}\:+\:\mathrm{16}} \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{40x}^{\mathrm{2}} \:+\:\mathrm{32x}\:+\:\mathrm{16} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{8x}^{\mathrm{3}} +\mathrm{16x}^{\mathrm{2}} −\mathrm{40x}^{\mathrm{2}} \:−\:\mathrm{32x}\:−\:\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{8x}^{\mathrm{3}} −\mathrm{24x}^{\mathrm{2}} \:−\:\mathrm{32x}\:−\:\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{3}.\mathrm{1850},−\mathrm{10}.\mathrm{0814}\:\left({Using}\:{calculator}\right) \\ $$$${You}\:{can}\:{see}\:{exact}\:{results}\:{in}\:{an} \\ $$$${advanced}\:{calculator}\:{by}\:{yourself}. \\ $$$${They}'{re}\:{too}\:{complicated}! \\ $$
Answered by MJS_new last updated on 16/Oct/21
x^2 +4x=(√(40x^2 +32x−16)) squaring & transforming x^4 +8x^3 −24x^2 −32x+16=0 (x^2 +4(1−(√2))x−4)(x^2 +4(1+(√2))x−4)=0 of the 4 solutions 3 fit the given equation x_1 =−2+2(√2)+2(√(4−2(√2))) x_(2, 3) =−2−2(√2)±2(√(4+2(√2)))
$${x}^{\mathrm{2}} +\mathrm{4}{x}=\sqrt{\mathrm{40}{x}^{\mathrm{2}} +\mathrm{32}{x}−\mathrm{16}} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming} \\ $$$${x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{3}} −\mathrm{24}{x}^{\mathrm{2}} −\mathrm{32}{x}+\mathrm{16}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{3}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$$${x}_{\mathrm{1}} =−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} =−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\pm\mathrm{2}\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Commented by MJS_new last updated on 16/Oct/21
yes. if the equation for a has only 1 real solution (Cardano) or if it has 3 real solutions of trigonometric shape we cannot use them. we must hope for easy solutions ∈Q to be found within the factors of ±n
$$\mathrm{yes}.\:\mathrm{if}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{for}\:\mathfrak{a}\:\mathrm{has}\:\mathrm{only}\:\mathrm{1}\:\mathrm{real} \\ $$$$\mathrm{solution}\:\left(\mathrm{Cardano}\right)\:\mathrm{or}\:\mathrm{if}\:\mathrm{it}\:\mathrm{has}\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\mathrm{of}\:\mathrm{trigonometric}\:\mathrm{shape}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{use}\:\mathrm{them}. \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{hope}\:\mathrm{for}\:\mathrm{easy}\:\mathrm{solutions}\:\in\mathbb{Q}\:\mathrm{to}\:\mathrm{be} \\ $$$$\mathrm{found}\:\mathrm{within}\:\mathrm{the}\:\mathrm{factors}\:\mathrm{of}\:\pm{n} \\ $$
Commented by MathSh last updated on 16/Oct/21
Very nice dear Ser thanks
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thanks} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Oct/21
Thanks for precious knowledge!
$$\boldsymbol{\mathcal{T}{hanks}}\:{for}\:{precious}\:{knowledge}! \\ $$
Commented by MJS_new last updated on 16/Oct/21
you′re welcome
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Oct/21
Sir, how did you factor the equation?
$${Sir},\:{how}\:{did}\:{you}\:{factor}\:{the}\:{equation}? \\ $$
Commented by MJS_new last updated on 16/Oct/21
the method I use: x^4 +ax^3 +bx^2 +cx+d=0 let x=t−(a/4) to get t^4 +pt^2 +qt+r=0 we want to find α, β, γ to get (t^2 −αt−β)(t^2 +αt−γ)=0 t^4 −(α^2 +β+γ)t^2 +α(γ−β)t+βγ=0 ⇒ { ((p=−(α^2 +β+γ))),((q=α(γ−β))),((r=βγ)) :} solve the 1^(st) & 2^(nd) for β & γ inserting in the 3^(rd) & transforming we get α^6 +jα^4 +kα^2 +l=0 let α=(√(a−(j/3))) to get a^3 +ma+n=0 if this has an “easy to handle” solution we succeed, otherwise the given equation has no useable exact solution.
$$\mathrm{the}\:\mathrm{method}\:\mathrm{I}\:\mathrm{use}: \\ $$$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}={t}−\frac{{a}}{\mathrm{4}}\:\mathrm{to}\:\mathrm{get} \\ $$$${t}^{\mathrm{4}} +{pt}^{\mathrm{2}} +{qt}+{r}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{find}\:\alpha,\:\beta,\:\gamma\:\:\mathrm{to}\:\mathrm{get} \\ $$$$\left({t}^{\mathrm{2}} −\alpha{t}−\beta\right)\left({t}^{\mathrm{2}} +\alpha{t}−\gamma\right)=\mathrm{0} \\ $$$${t}^{\mathrm{4}} −\left(\alpha^{\mathrm{2}} +\beta+\gamma\right){t}^{\mathrm{2}} +\alpha\left(\gamma−\beta\right){t}+\beta\gamma=\mathrm{0} \\ $$$$\Rightarrow\:\begin{cases}{{p}=−\left(\alpha^{\mathrm{2}} +\beta+\gamma\right)}\\{{q}=\alpha\left(\gamma−\beta\right)}\\{{r}=\beta\gamma}\end{cases} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\&\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{for}\:\beta\:\&\:\gamma \\ $$$$\mathrm{inserting}\:\mathrm{in}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\&\:\mathrm{transforming}\:\mathrm{we}\:\mathrm{get} \\ $$$$\alpha^{\mathrm{6}} +{j}\alpha^{\mathrm{4}} +{k}\alpha^{\mathrm{2}} +{l}=\mathrm{0} \\ $$$$\mathrm{let}\:\alpha=\sqrt{\mathfrak{a}−\frac{{j}}{\mathrm{3}}}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathfrak{a}^{\mathrm{3}} +{m}\mathfrak{a}+{n}=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{this}\:\mathrm{has}\:\mathrm{an}\:“\mathrm{easy}\:\mathrm{to}\:\mathrm{handle}''\:\mathrm{solution}\:\mathrm{we} \\ $$$$\mathrm{succeed},\:\mathrm{otherwise}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{has} \\ $$$$\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solution}. \\ $$
Commented by MJS_new last updated on 16/Oct/21
x^2 +4x=(√(40x^2 +32x+16)) has no useable exact solution
$${x}^{\mathrm{2}} +\mathrm{4}{x}=\sqrt{\mathrm{40}{x}^{\mathrm{2}} +\mathrm{32}{x}+\mathrm{16}}\:\mathrm{has}\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solution} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Oct/21
Thanks a lot sir! I used a calculator and it gave me terrible answers in exact form!
$$\boldsymbol{\mathcal{T}{hanks}}\:\boldsymbol{{a}}\:\boldsymbol{{lot}}\:\boldsymbol{{sir}}!\:{I}\:{used}\:{a}\:{calculator} \\ $$$${and}\:{it}\:{gave}\:{me}\:{terrible}\:{answers} \\ $$$${in}\:{exact}\:{form}! \\ $$

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