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Question-221669

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Question Number 221669 by Tawa11 last updated on 09/Jun/25
Commented by Tawa11 last updated on 09/Jun/25
Solve for t.
$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{t}. \\ $$
Commented by AlagaIbile last updated on 09/Jun/25
Commented by Tawa11 last updated on 09/Jun/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by fantastic last updated on 09/Jun/25
((t^7 +t^5 +t^3 )/(t^6 +t^5 +t^4 ))=((81)/(27)) or ((t^3 (t^4 +t^2 +1))/(t^4 (t^2 +t+1)))=3 or (((t^2 )^2 +2.t^2 +(1)^2 −t^2 )/(t(t^2 +t+1)))=3 or (((t^2 +1)^2 −(t)^2 )/(t(t^2 +t+1)))=3 or (((t^2 +1+t)(t^2 +1−t))/(t(t^2 +t+1)))=3 or t^2 −t+1 =3t or t^2 −4t+1=0 So t=((−(−4)±(√((−4)^2 −4(1)(1))))/(2(1)))=((4±(√(16−4)))/2)=2±((√(12))/2)=2±((2(√3))/2)=2±(√3) So t=2+(√3)∨2−(√3)
$$\frac{{t}^{\mathrm{7}} +{t}^{\mathrm{5}} +{t}^{\mathrm{3}} }{{t}^{\mathrm{6}} +{t}^{\mathrm{5}} +{t}^{\mathrm{4}} }=\frac{\mathrm{81}}{\mathrm{27}} \\ $$$${or}\:\frac{{t}^{\mathrm{3}} \left({t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}\right)}{{t}^{\mathrm{4}} \left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}=\mathrm{3} \\ $$$${or}\:\frac{\left({t}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}.{t}^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} −{t}^{\mathrm{2}} }{{t}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}=\mathrm{3} \\ $$$${or}\:\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left({t}\right)^{\mathrm{2}} }{{t}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}=\mathrm{3} \\ $$$${or}\:\frac{\cancel{\left({t}^{\mathrm{2}} +\mathrm{1}+{t}\right)}\left({t}^{\mathrm{2}} +\mathrm{1}−{t}\right)}{{t}\cancel{\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}}=\mathrm{3} \\ $$$${or}\:{t}^{\mathrm{2}} −{t}+\mathrm{1}\:=\mathrm{3}{t} \\ $$$${or}\:{t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{1}=\mathrm{0} \\ $$$${So}\:{t}=\frac{−\left(−\mathrm{4}\right)\pm\sqrt{\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right)}}{\mathrm{2}\left(\mathrm{1}\right)}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}}{\mathrm{2}}=\mathrm{2}\pm\frac{\sqrt{\mathrm{12}}}{\mathrm{2}}=\mathrm{2}\pm\frac{\cancel{\mathrm{2}}\sqrt{\mathrm{3}}}{\cancel{\mathrm{2}}}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$${So}\:\underline{{t}=\mathrm{2}+\sqrt{\mathrm{3}}}\vee\underline{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$
Commented by Tawa11 last updated on 09/Jun/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by Rasheed.Sindhi last updated on 09/Jun/25
((t^7 +t^5 +t^3 )/(t^6 +t^5 +t^4 ))=((81)/(27)) ((t^5 (t^2 +1+(1/t^2 )))/(t^5 (t+1+(1/t))))=3 ;t≠0 ((t^2 +1+(1/t^2 ))/(t+1+(1/t)))=3 (((t+(1/t))^2 −1)/((t+(1/t)+1)))=3 (((t+(1/t)+1)(t+(1/t)−1))/((t+(1/t)+1)))=3 t+(1/t)−4=0 t^2 −4t+1=0 t=((4±(√(16−4)))/2) t=((4±2(√3))/2)=2±(√3)
$$\frac{\mathrm{t}^{\mathrm{7}} +\mathrm{t}^{\mathrm{5}} +\mathrm{t}^{\mathrm{3}} }{\mathrm{t}^{\mathrm{6}} +\mathrm{t}^{\mathrm{5}} +\mathrm{t}^{\mathrm{4}} }=\frac{\mathrm{81}}{\mathrm{27}} \\ $$$$\frac{\mathrm{t}^{\mathrm{5}} \left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}{\mathrm{t}^{\mathrm{5}} \left(\mathrm{t}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}}\right)}=\mathrm{3}\:\:;\mathrm{t}\neq\mathrm{0} \\ $$$$\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}}}=\mathrm{3}\:\: \\ $$$$\frac{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}\right)}=\mathrm{3} \\ $$$$\frac{\cancel{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}\right)}\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}−\mathrm{1}\right)}{\cancel{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}\right)}}=\mathrm{3} \\ $$$$\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{t}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{t}=\frac{\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}}\: \\ $$
Commented by Tawa11 last updated on 09/Jun/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by Rasheed.Sindhi last updated on 10/Jun/25
(((t+(1/t)+1)(t+(1/t)−1))/((t+(1/t)+1)))=3 (t+(1/t)+1)(t+(1/t)−1)−3(t+(1/t)+1)=0 (t+(1/t)+1)((t+(1/t)−1)−3)=0 •t+(1/t)+1=0⇒t^2 +t+1=0 t=((−1±(√(1−4)))/2)=((−1±i(√3) )/2) •(t+(1/t)−1)−3=0⇒t^2 −4t+1=0 t=((4±(√(16−4)) )/2)=2±(√3)
$$\frac{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}\right)\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}−\mathrm{1}\right)}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}\right)}=\mathrm{3} \\ $$$$\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}\right)\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}−\mathrm{1}\right)−\mathrm{3}\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}\right)\left(\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}−\mathrm{1}\right)−\mathrm{3}\right)=\mathrm{0} \\ $$$$\bullet\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{t}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}}=\frac{−\mathrm{1}\pm\mathrm{i}\sqrt{\mathrm{3}}\:}{\mathrm{2}} \\ $$$$\bullet\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}−\mathrm{1}\right)−\mathrm{3}=\mathrm{0}\Rightarrow\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{t}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}\:}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}}\: \\ $$
Commented by Tawa11 last updated on 10/Jun/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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