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Question Number 68835 by MJS last updated on 16/Sep/19
reposting this because it′s not yet solved correctly. I criticize that most of you people do not test if your solutions fit the given equations in many cases x^2 +1−(√(x^3 +x))=6x please determine (1) how many real solutions (2) how many complex solutions we can expect (3) solve it (4) test your solutions
$$\mathrm{reposting}\:\mathrm{this}\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{yet}\:\mathrm{solved} \\ $$$$\mathrm{correctly}.\:\mathrm{I}\:\mathrm{criticize}\:\mathrm{that}\:\mathrm{most}\:\mathrm{of}\:\mathrm{you}\:\mathrm{people} \\ $$$$\mathrm{do}\:\mathrm{not}\:\mathrm{test}\:\mathrm{if}\:\mathrm{your}\:\mathrm{solutions}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equations}\:\mathrm{in}\:\mathrm{many}\:\mathrm{cases} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\mathrm{1}−\sqrt{{x}^{\mathrm{3}} +{x}}=\mathrm{6}{x} \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{determine} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{how}\:\mathrm{many}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{how}\:\mathrm{many}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{expect} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{solve}\:\mathrm{it} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{test}\:\mathrm{your}\:\mathrm{solutions} \\ $$
Commented by Prithwish sen last updated on 16/Sep/19
x^2 −6x+1=(√(x^3 +x)) Squaring both sides we get x^4 −13x^3 +38x^2 −13x+1=0 and this is the reciprocal equation of 1^(st) type ∴ (x^4 +1)−13x(x^2 +1)+38x^2 =0 Dividing the equation by x^2 we get (x^2 +(1/x^2 ))−13(x+(1/x))+38 =0 (x+(1/x))^2 −13(x+(1/x))+36=0 putting x+(1/x) = y we get y^2 −13y+36 =0 (y−4)(y−9)=0 and i.e x^2 −4x+1= 0 and x^2 −9x +1= 0 Solving we get x = 2± (√3) , ((9±(√(77)))/2) now 2−(√3) and 2+(√3) are reciprocal on the other hand ((9+(√(77)))/2) and ((9−(√(77)))/2) are reciprocal. now 2+(√(3 ))∽ 3.73 and 2−(√3) ∽ 0.268 ((9−(√(77)))/2) ∽ 0.11 and ((9+(√(77)))/2) ∽ 8.89 the equation satisfies only on 0.11 and 8.89 MJS Sir, please share your valuable comments.
$$\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{1}=\sqrt{\mathrm{x}^{\mathrm{3}} +\mathrm{x}} \\ $$$$\mathrm{Squaring}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{we}\:\mathrm{get} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\mathrm{13}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{38}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{13}\boldsymbol{\mathrm{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{reciprocal}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{of}}\:\mathrm{1}^{\boldsymbol{\mathrm{st}}} \:\boldsymbol{\mathrm{type}} \\ $$$$\:\therefore\:\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)−\mathrm{13x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{38x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{Dividing}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{by}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)−\mathrm{13}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{38}\:=\mathrm{0} \\ $$$$\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} −\mathrm{13}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{36}=\mathrm{0} \\ $$$$\mathrm{putting}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\:=\:\boldsymbol{\mathrm{y}}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{13y}+\mathrm{36}\:=\mathrm{0} \\ $$$$\left(\mathrm{y}−\mathrm{4}\right)\left(\mathrm{y}−\mathrm{9}\right)=\mathrm{0}\:\:\:\mathrm{and}\:\mathrm{i}.\mathrm{e} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}=\:\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{9x}\:+\mathrm{1}=\:\mathrm{0} \\ $$$$\mathrm{Solving}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\:\:\boldsymbol{\mathrm{x}}\:=\:\mathrm{2}\pm\:\sqrt{\mathrm{3}}\:\:,\:\frac{\mathrm{9}\pm\sqrt{\mathrm{77}}}{\mathrm{2}}\:\: \\ $$$$\boldsymbol{\mathrm{now}}\:\mathrm{2}−\sqrt{\mathrm{3}}\:\boldsymbol{\mathrm{and}}\:\mathrm{2}+\sqrt{\mathrm{3}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{reciprocal}} \\ $$$$\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{other}}\:\boldsymbol{\mathrm{hand}}\:\frac{\mathrm{9}+\sqrt{\mathrm{77}}}{\mathrm{2}}\:\:\boldsymbol{\mathrm{and}}\:\frac{\mathrm{9}−\sqrt{\mathrm{77}}}{\mathrm{2}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{reciprocal}}. \\ $$$$\boldsymbol{\mathrm{now}}\:\mathrm{2}+\sqrt{\mathrm{3}\:}\backsim\:\mathrm{3}.\mathrm{73}\:\:\boldsymbol{\mathrm{and}}\:\mathrm{2}−\sqrt{\mathrm{3}}\:\backsim\:\mathrm{0}.\mathrm{268} \\ $$$$\frac{\mathrm{9}−\sqrt{\mathrm{77}}}{\mathrm{2}}\:\backsim\:\mathrm{0}.\mathrm{11}\:\:\boldsymbol{\mathrm{and}}\:\frac{\mathrm{9}+\sqrt{\mathrm{77}}}{\mathrm{2}}\:\backsim\:\mathrm{8}.\mathrm{89} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{satisfies}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{on}}\:\mathrm{0}.\mathrm{11}\:\boldsymbol{\mathrm{and}}\:\mathrm{8}.\mathrm{89} \\ $$$$\boldsymbol{\mathrm{MJS}}\:\boldsymbol{\mathrm{Sir}},\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{share}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{valuable}}\:\boldsymbol{\mathrm{comments}}. \\ $$
Commented by MJS last updated on 16/Sep/19
good, thank you! we have to test, because with squaring we get false solutions
$$\mathrm{good},\:\mathrm{thank}\:\mathrm{you}!\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{test},\:\mathrm{because} \\ $$$$\mathrm{with}\:\mathrm{squaring}\:\mathrm{we}\:\mathrm{get}\:\mathrm{false}\:\mathrm{solutions} \\ $$
Commented by Prithwish sen last updated on 16/Sep/19
Thank you sir.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Rasheed.Sindhi last updated on 16/Sep/19
Slightly different(?) way x^2 +1−(√(x^3 +x))=6x x(x+(1/x))−x(√(x+(1/x)))−6x=0 x{(x+(1/x))−(√(x+(1/x)))−6}=0 x=0 (Not valid) ∣ (x+(1/x))−(√(x+(1/x)))−6=0 Let (√(x+(1/x)))=y y^2 −y−6=0 (y−3)(y+2)=0 y=3 ∣ y=−2 (√(x+(1/x)))=3 ∣ (√(x+(1/x)))=−2 (false) x+(1/x)=9 x^2 −9x+1=0 x=((9±(√(81−4)))/2)=((9±(√(77)))/2) Substituting in original equation we can see that these both are valid.
$$\mathrm{Slightly}\:\mathrm{different}\left(?\right)\:\mathrm{way} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{x}^{\mathrm{3}} +\mathrm{x}}=\mathrm{6x} \\ $$$$\mathrm{x}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{x}\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}}−\mathrm{6x}=\mathrm{0} \\ $$$$\mathrm{x}\left\{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)−\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}}−\mathrm{6}\right\}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{0}\:\left(\mathrm{Not}\:\mathrm{valid}\right)\:\mid\:\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)−\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}}−\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{Let}\:\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}}=\mathrm{y} \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{y}−\mathrm{6}=\mathrm{0} \\ $$$$\left(\mathrm{y}−\mathrm{3}\right)\left(\mathrm{y}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{3}\:\:\mid\:\:\:\mathrm{y}=−\mathrm{2} \\ $$$$\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}}=\mathrm{3}\:\:\mid\:\:\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}}=−\mathrm{2}\:\left(\mathrm{false}\right) \\ $$$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{9} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{9x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{9}\pm\sqrt{\mathrm{77}}}{\mathrm{2}} \\ $$$$\mathrm{Substituting}\:\mathrm{in}\:\mathrm{original}\:\mathrm{equation}\:\mathrm{we} \\ $$$$\mathrm{can}\:\mathrm{see}\:\mathrm{that}\:\mathrm{these}\:\mathrm{both}\:\mathrm{are}\:\mathrm{valid}. \\ $$
Answered by ajfour last updated on 16/Sep/19
(x^2 +1−6x)^2 =x^3 +x dividing by x^2 (x+(1/x)−6)^2 =x+(1/x) if we let x+(1/x)=t t^2 −13t+36=0 t=((13)/2)±(√(((169)/4)−36)) t=((13)/2)±(5/2) ⇒ x+(1/x)=4, 9 x^2 −4x+1=0 ⇒ x=2±(√(4−1)) = 2±(√3) >0 and from x^2 −9x+1=0 x=(9/2)±(√(((81)/4)−1)) =(9/2)±((√(77))/2) >0 x^3 −4x^2 +x=0 ⇒ x^3 +x=4x^2 or x^3 −9x^2 +x=0 ⇒ x^3 +x=9x^2 testing x^2 +1−6x=4∣x∣ and as all found roots are >0 ⇒ x^2 −10x+1=0 ⇒ x=5±(√(25−1)) = 5±2(√6) ⇒ x=2±(√3) cannot be true solutions. x^2 +1−6x=3∣x∣ gives for x>0 x^2 −9x+1=0 (same eq.) ⇒so x=((9±(√(77)))/2) are true solutions.
$$\left({x}^{\mathrm{2}} +\mathrm{1}−\mathrm{6}{x}\right)^{\mathrm{2}} ={x}^{\mathrm{3}} +{x} \\ $$$${dividing}\:{by}\:{x}^{\mathrm{2}} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}−\mathrm{6}\right)^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{{x}} \\ $$$${if}\:{we}\:{let}\:\:{x}+\frac{\mathrm{1}}{{x}}={t} \\ $$$$\:\:\:{t}^{\mathrm{2}} −\mathrm{13}{t}+\mathrm{36}=\mathrm{0} \\ $$$$\:\:\:{t}=\frac{\mathrm{13}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{169}}{\mathrm{4}}−\mathrm{36}} \\ $$$$\:\:\:{t}=\frac{\mathrm{13}}{\mathrm{2}}\pm\frac{\mathrm{5}}{\mathrm{2}}\: \\ $$$$\Rightarrow\:\:\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{4},\:\mathrm{9} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}=\mathrm{0}\:\:\Rightarrow\:\: \\ $$$$\:\:\:\:\:{x}=\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{1}}\:=\:\mathrm{2}\pm\sqrt{\mathrm{3}}\:\:>\mathrm{0} \\ $$$${and}\:{from}\:\:{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\frac{\mathrm{9}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{81}}{\mathrm{4}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{9}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{77}}}{\mathrm{2}}\:\:\:\:>\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}=\mathrm{0}\:\:\:\Rightarrow\:{x}^{\mathrm{3}} +{x}=\mathrm{4}{x}^{\mathrm{2}} \\ $$$${or}\:\:{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +{x}=\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{x}=\mathrm{9}{x}^{\mathrm{2}} \\ $$$${testing} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{6}{x}=\mathrm{4}\mid{x}\mid\:\:\:{and}\:{as}\:{all} \\ $$$$\:\:{found}\:{roots}\:{are}\:>\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:{x}=\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{1}}\:=\:\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\:\:\:\:\Rightarrow\:\:{x}=\mathrm{2}\pm\sqrt{\mathrm{3}}\:\:{cannot}\:{be}\:{true} \\ $$$$\:\:\:\:{solutions}. \\ $$$$\:\:\:{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{6}{x}=\mathrm{3}\mid{x}\mid\:\:\:{gives}\:{for}\:{x}>\mathrm{0} \\ $$$$\:\:\:{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{1}=\mathrm{0}\:\:\left({same}\:{eq}.\right) \\ $$$$\Rightarrow{so}\:{x}=\frac{\mathrm{9}\pm\sqrt{\mathrm{77}}}{\mathrm{2}}\:{are}\:{true}\:{solutions}. \\ $$
Commented by MJS last updated on 16/Sep/19
good, thank you!
$$\mathrm{good},\:\mathrm{thank}\:\mathrm{you}! \\ $$

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