Question-195175

Question Number 195175 by valdirmd last updated on 25/Jul/23

Answered by Rasheed.Sindhi last updated on 26/Jul/23

x^4 +x^2 =11/5 , (((x+1)/(x−1)))^(1/3) +(((x−1)/(x+1)))^(1/3) =? ▶Let y=(((x+1)/(x−1)))^(1/3) +(((x−1)/(x+1)))^(1/3) y^3 =((x+1)/(x−1))+((x−1)/(x+1))+3y y^3 −3y=((x^2 +2x+1+x^2 −2x+1)/(x^2 −1)) y^3 −3y=2(((x^2 −1+2)/(x^2 −1)))=2(1+(2/(x^2 −1))) =2+(4/(x^2 −1)) ▶ x^4 +x^2 =11/5⇒5x^4 +5x^2 −11=0 x^2 (≥0)=((−5+(√(25+220)))/(10))=((−5+7(√5))/(10)) ▶y^3 −3y=2+(4/(x^2 −1))=2+(4/(((−5+7(√5))/(10))−1)) =2+(4/((−15+7(√5) )/(10)))=2+((40)/(−15+7(√5) )) =2+((40(−15−7(√5) ))/(225−245)) y^3 −3y =2−2(−15−7(√5) ) =2+30+14(√5) = y^3 −3y=32+14(√5) ▶(((x+1)/(x−1)))^(1/3) +(((x−1)/(x+1)))^(1/3) =y=3.7361626

$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} =\mathrm{11}/\mathrm{5}\:,\:\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} =? \\ $$$$\blacktriangleright{Let}\:{y}=\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:{y}^{\mathrm{3}} =\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}+\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}+\mathrm{3}{y} \\ $$$$\:\:\:\:\:{y}^{\mathrm{3}} −\mathrm{3}{y}=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:\:\:{y}^{\mathrm{3}} −\mathrm{3}{y}=\mathrm{2}\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}}\right)=\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\blacktriangleright\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} =\mathrm{11}/\mathrm{5}\Rightarrow\mathrm{5}{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{11}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} \left(\geqslant\mathrm{0}\right)=\frac{−\mathrm{5}+\sqrt{\mathrm{25}+\mathrm{220}}}{\mathrm{10}}=\frac{−\mathrm{5}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$\blacktriangleright{y}^{\mathrm{3}} −\mathrm{3}{y}=\mathrm{2}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2}+\frac{\mathrm{4}}{\frac{−\mathrm{5}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}+\frac{\mathrm{4}}{\frac{−\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}\:}{\mathrm{10}}}=\mathrm{2}+\frac{\mathrm{40}}{−\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}+\frac{\mathrm{40}\left(−\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}\:\right)}{\mathrm{225}−\mathrm{245}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{3}} −\mathrm{3}{y}\:=\mathrm{2}−\mathrm{2}\left(−\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}+\mathrm{30}+\mathrm{14}\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{3}} −\mathrm{3}{y}=\mathrm{32}+\mathrm{14}\sqrt{\mathrm{5}} \\ $$$$\blacktriangleright\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} ={y}=\mathrm{3}.\mathrm{7361626} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Jul/23

ThanX sir! I′m going to correct my answer.

$$\mathcal{T}{han}\mathcal{X}\:{sir}!\:{I}'{m}\:{going}\:{to}\:{correct}\:{my} \\ $$$${answer}. \\ $$

Commented by jihamine2301 last updated on 26/Jul/23

Error at y^3 =.....+3y and x^2 should get only positive value

$${Error}\:\:{at}\:\:{y}^{\mathrm{3}} =…..+\mathrm{3}{y}\:{and}\:{x}^{\mathrm{2}} \:{should}\:{get} \\ $$$${only}\:{positive}\:{value} \\ $$

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