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Question Number 145114 by imjagoll last updated on 02/Jul/21
Commented by MJS_new last updated on 02/Jul/21
for x∈R I get 2+(√5)
$$\mathrm{for}\:{x}\in\mathbb{R}\:\mathrm{I}\:\mathrm{get}\:\mathrm{2}+\sqrt{\mathrm{5}} \\ $$
Commented by imjagoll last updated on 02/Jul/21
how did the answer sir?
$$\mathrm{how}\:\mathrm{did}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{sir}? \\ $$
Commented by Canebulok last updated on 02/Jul/21
Random Problem: If x^4 +x^2 = ((11)/5) What is the value of ^3 (√((x+1)/(x−1))) +^3 (√((x−1)/(x+1))) = ? Solution: ⇒ 5x^4 +5x^2 −11 = 0 let: ⇒ x^2 = K ∵ ⇒ 5K^2 +5K−11 = 0 By quadratic formula, ⇒ K = ((−5±(√(5^2 −4(5)(−11))))/(2(5))) ⇒ K_1 = ((−5+7(√5))/(10)) ⇒ K_2 = ((−5−7(√5))/(10)) ∴ ⇒ x^2 = ((−5±7(√5))/(10)) ⇒ x = (√((((−5±7(√5))/(10))))) Thus; ⇒^3 (√((x+1)/(x−1))) +^3 (√((x−1)/(x+1))) ≈ 1.837977 or ⇒^3 (√((x+1)/(x−1))) +^3 (√((x−1)/(x+1))) ≈ 4.236068
$$\boldsymbol{{Random}}\:\boldsymbol{{Problem}}: \\ $$$$\:{If}\:\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} \:=\:\frac{\mathrm{11}}{\mathrm{5}} \\ $$$$\:{What}\:{is}\:{the}\:{value}\:{of} \\ $$$$\:\:^{\mathrm{3}} \sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\:^{\mathrm{3}} \sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:=\:? \\ $$$$\boldsymbol{{Solution}}: \\ $$$$\Rightarrow\:\mathrm{5}{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{11}\:=\:\mathrm{0} \\ $$$${let}: \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:=\:{K} \\ $$$$\because \\ $$$$\Rightarrow\:\mathrm{5}{K}^{\mathrm{2}} +\mathrm{5}{K}−\mathrm{11}\:=\:\mathrm{0} \\ $$$${By}\:{quadratic}\:{formula}, \\ $$$$\Rightarrow\:{K}\:=\:\frac{−\mathrm{5}\pm\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{5}\right)\left(−\mathrm{11}\right)}}{\mathrm{2}\left(\mathrm{5}\right)} \\ $$$$\Rightarrow\:{K}_{\mathrm{1}} \:=\:\frac{−\mathrm{5}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$\Rightarrow\:{K}_{\mathrm{2}} \:=\:\frac{−\mathrm{5}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$\therefore \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:=\:\frac{−\mathrm{5}\pm\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$\Rightarrow\:{x}\:=\:\sqrt{\left(\frac{−\mathrm{5}\pm\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)} \\ $$$${Thus}; \\ $$$$\Rightarrow\:^{\mathrm{3}} \sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\:^{\mathrm{3}} \sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:\:\approx\:\mathrm{1}.\mathrm{837977} \\ $$$${or} \\ $$$$\Rightarrow\:^{\mathrm{3}} \sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\:^{\mathrm{3}} \sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:\:\approx\:\mathrm{4}.\mathrm{236068} \\ $$$$\: \\ $$
Commented by Canebulok last updated on 02/Jul/21
sorry i didn't notice
Answered by liberty last updated on 02/Jul/21
⇒ 5x^4 +5x^2 −11=0 ⇒x^2 = ((−5+(√(25+220)))/(10)) ⇒x^2 +1 = ((5+(√(245)))/(10)) ⇒x^2 +1=((5+7(√5))/(10))...(i) ⇒x^2 −1=((−15+7(√5))/(10))...(ii) ⇒((x^2 +1)/(x^2 −1))=((5+7(√5))/(−15+7(√5)))...(iii) consider (((x+1)/(x−1)))^(1/3) +(((x−1)/(x+1)))^(1/3) = ((((x+1)/(x−1)) +((x−1)/(x+1)))/( (((((x+1)/(x−1)))^2 ))^(1/3) + (((((x−1)/(x+1)))^2 ))^(1/3) −1)) = (((2(x^2 +1))/(x^2 −1))/( (((((x+1)^3 )/((x^2 −1)^2 )) ))^(1/3) +((((x−1)^3 )/((x^2 −1)^2 )))^(1/3) −1)) = ((2(((7(√5)+5)/(7(√5)−15))))/( ((2x)/( (((((7(√5)−15)/(10)))^2 ))^(1/3) )) −1)) ...
$$\Rightarrow\:\mathrm{5}{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{11}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \:=\:\frac{−\mathrm{5}+\sqrt{\mathrm{25}+\mathrm{220}}}{\mathrm{10}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{1}\:=\:\frac{\mathrm{5}+\sqrt{\mathrm{245}}}{\mathrm{10}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{5}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}…\left({i}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{1}=\frac{−\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}…\left({ii}\right) \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{5}+\mathrm{7}\sqrt{\mathrm{5}}}{−\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}…\left({iii}\right) \\ $$$${consider}\:\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\: \\ $$$$=\:\frac{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\:+\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}{\:\sqrt[{\mathrm{3}}]{\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} }\:+\:\sqrt[{\mathrm{3}}]{\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} }−\mathrm{1}} \\ $$$$=\:\frac{\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt[{\mathrm{3}}]{\frac{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:}\:+\sqrt[{\mathrm{3}}]{\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{2}\left(\frac{\mathrm{7}\sqrt{\mathrm{5}}+\mathrm{5}}{\mathrm{7}\sqrt{\mathrm{5}}−\mathrm{15}}\right)}{\:\frac{\mathrm{2}{x}}{\:\sqrt[{\mathrm{3}}]{\left(\frac{\mathrm{7}\sqrt{\mathrm{5}}−\mathrm{15}}{\mathrm{10}}\right)^{\mathrm{2}} }}\:−\mathrm{1}} \\ $$$$\:… \\ $$
Answered by MJS_new last updated on 02/Jul/21
a^(1/3) +b^(1/3) =c a+3a^(1/3) b^(1/3) (a^(1/3) +b^(1/3) )+b=c^3 a+3a^(1/3) b^(1/3) c+b=c^3 3a^(1/3) b^(1/3) c=c^3 −a−b 27abc^3 =(c^3 −a−b)^3 c^9 −3(a+b)c^6 +3(a^2 −7ab+b^2 )c^3 −(a+b)^3 =0 a=((x+1)/(x−1))∧b=(1/a) c^9 −((6(x^2 +1))/(x^2 −1))c^6 −((3(x^2 −5)(5x^2 −1))/((x^2 −1)^2 ))c^3 −((8(x^2 +1)^3 )/((x^2 −1)^3 ))=0 x^4 +x^2 =((11)/5) ⇒ x^2 =−(1/2)+((7(√5))/(10)) c^9 −6(16+7(√5))c^6 +21(285+128(√5))c^3 −8(15856+7091(√5))=0 c=(z+2(16+7(√5)))^(1/3) z^3 −27z−54(16+7(√5))=0 (z−3(2+(√5)))(z^2 +3(2+(√5))z+18(3+2(√5)))=0 ⇒ z=3(2+(√5)) ⇒ c=((38+17(√5)))^(1/3) =2+(√5)
$${a}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{1}/\mathrm{3}} ={c} \\ $$$${a}+\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} \left({a}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{1}/\mathrm{3}} \right)+{b}={c}^{\mathrm{3}} \\ $$$${a}+\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} {c}+{b}={c}^{\mathrm{3}} \\ $$$$\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} {c}={c}^{\mathrm{3}} −{a}−{b} \\ $$$$\mathrm{27}{abc}^{\mathrm{3}} =\left({c}^{\mathrm{3}} −{a}−{b}\right)^{\mathrm{3}} \\ $$$${c}^{\mathrm{9}} −\mathrm{3}\left({a}+{b}\right){c}^{\mathrm{6}} +\mathrm{3}\left({a}^{\mathrm{2}} −\mathrm{7}{ab}+{b}^{\mathrm{2}} \right){c}^{\mathrm{3}} −\left({a}+{b}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$${a}=\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\wedge{b}=\frac{\mathrm{1}}{{a}} \\ $$$${c}^{\mathrm{9}} −\frac{\mathrm{6}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}{c}^{\mathrm{6}} −\frac{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{5}\right)\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{c}^{\mathrm{3}} −\frac{\mathrm{8}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} =\frac{\mathrm{11}}{\mathrm{5}}\:\Rightarrow\:{x}^{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$${c}^{\mathrm{9}} −\mathrm{6}\left(\mathrm{16}+\mathrm{7}\sqrt{\mathrm{5}}\right){c}^{\mathrm{6}} +\mathrm{21}\left(\mathrm{285}+\mathrm{128}\sqrt{\mathrm{5}}\right){c}^{\mathrm{3}} −\mathrm{8}\left(\mathrm{15856}+\mathrm{7091}\sqrt{\mathrm{5}}\right)=\mathrm{0} \\ $$$${c}=\left({z}+\mathrm{2}\left(\mathrm{16}+\mathrm{7}\sqrt{\mathrm{5}}\right)\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${z}^{\mathrm{3}} −\mathrm{27}{z}−\mathrm{54}\left(\mathrm{16}+\mathrm{7}\sqrt{\mathrm{5}}\right)=\mathrm{0} \\ $$$$\left({z}−\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\right)\left({z}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right){z}+\mathrm{18}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${z}=\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:\Rightarrow\:{c}=\sqrt[{\mathrm{3}}]{\mathrm{38}+\mathrm{17}\sqrt{\mathrm{5}}}=\mathrm{2}+\sqrt{\mathrm{5}} \\ $$
Commented by MJS_new last updated on 02/Jul/21
is this really so hard? I can′t believe...
$$\mathrm{is}\:\mathrm{this}\:\mathrm{really}\:\mathrm{so}\:\mathrm{hard}?\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{believe}… \\ $$
Commented by imjagoll last updated on 03/Jul/21
hahaha i believe it problem super hard sir
$$\mathrm{hahaha}\:\mathrm{i}\:\mathrm{believe}\:\mathrm{it}\:\mathrm{problem}\:\mathrm{super}\: \\ $$$$\mathrm{hard}\:\mathrm{sir} \\ $$

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