Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 149766 by mathdanisur last updated on 07/Aug/21

Find the roots of the equation:  x^2  + x + 1 + (1/(x^2  + x + 1)) = ((10)/3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}}\:=\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$

Commented by amin96 last updated on 07/Aug/21

x^2 +x+1=y    ⇒  y+(1/y)=((10)/3)  y^2 +1=((10y)/3)  ⇒  3y^2 −10y+3=0  Δ=100−36=64    y=((10+8)/6)=3  x^2 +x+1=3  ⇒  x^2 +x−2=0   Δ=9   { ((x_1 =((−1+3)/2)=1)),((x_2 =((−1−3)/2)=−2)) :}

$${x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{y}\:\:\:\:\Rightarrow\:\:{y}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$$${y}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{10}{y}}{\mathrm{3}}\:\:\Rightarrow\:\:\mathrm{3}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{3}=\mathrm{0} \\ $$$$\Delta=\mathrm{100}−\mathrm{36}=\mathrm{64}\:\:\:\:{y}=\frac{\mathrm{10}+\mathrm{8}}{\mathrm{6}}=\mathrm{3} \\ $$$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{3}\:\:\Rightarrow\:\:{x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0}\:\:\:\Delta=\mathrm{9} \\ $$$$\begin{cases}{{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{3}}{\mathrm{2}}=\mathrm{1}}\\{{x}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{3}}{\mathrm{2}}=−\mathrm{2}}\end{cases}\:\: \\ $$

Commented by mathdanisur last updated on 07/Aug/21

Thankyou Ser

$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$

Answered by Canebulok last updated on 07/Aug/21

   Solution:  let:  ⇒ x^2 +x+1 = k  ∵  ⇒ k+(1/k) = ((10)/3)  ⇒ k^2 +1 = ((10)/3)∙k  ⇒ k^2 −((10)/3)∙k+1 = 0  By using quadratic formula:  ⇒ k = (((((10)/3))±(√((−((10)/3))^2 −4)))/2)               = (((((10)/3))±(√((64)/9)))/2)  ∵  ⇒ k = 3   ;   k = (1/3)  By substituting back:  (1st solutions)  ⇒ x^2 +x+1 = 3  ⇒ x^2 +x−2 = 0  ⇒ (x+2)(x−1) = 0  ⇒ x_1  = −2   ;  x_2  = 1     (2nd solutions)  ⇒ x^2 +x+1 = (1/3)  ⇒ x^2 +x+(2/3) = 0  By quadratic formula:  ⇒ x = ((−1±(√(1^2 −4((2/3)))))/2)           x_(3,4)   = ((−1±i(√(5/3)))/2)     Solution by: Kevin

$$\: \\ $$$$\boldsymbol{{Solution}}: \\ $$$${let}: \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:{k} \\ $$$$\because \\ $$$$\Rightarrow\:{k}+\frac{\mathrm{1}}{{k}}\:=\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$$$\Rightarrow\:{k}^{\mathrm{2}} +\mathrm{1}\:=\:\frac{\mathrm{10}}{\mathrm{3}}\centerdot{k} \\ $$$$\Rightarrow\:{k}^{\mathrm{2}} −\frac{\mathrm{10}}{\mathrm{3}}\centerdot{k}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${By}\:{using}\:{quadratic}\:{formula}: \\ $$$$\Rightarrow\:{k}\:=\:\frac{\left(\frac{\mathrm{10}}{\mathrm{3}}\right)\pm\sqrt{\left(−\frac{\mathrm{10}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\frac{\mathrm{10}}{\mathrm{3}}\right)\pm\sqrt{\frac{\mathrm{64}}{\mathrm{9}}}}{\mathrm{2}} \\ $$$$\because \\ $$$$\Rightarrow\:{k}\:=\:\mathrm{3}\:\:\:;\:\:\:{k}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${By}\:{substituting}\:{back}: \\ $$$$\left(\mathrm{1}{st}\:{solutions}\right) \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:\mathrm{3} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{2}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\left({x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} \:=\:−\mathrm{2}\:\:\:;\:\:{x}_{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\: \\ $$$$\left(\mathrm{2}{nd}\:{solutions}\right) \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{x}+\frac{\mathrm{2}}{\mathrm{3}}\:=\:\mathrm{0} \\ $$$${By}\:{quadratic}\:{formula}: \\ $$$$\Rightarrow\:{x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}}{\mathrm{2}} \\ $$$$\: \\ $$$$\:\:\:\:\:\:{x}_{\mathrm{3},\mathrm{4}} \:\:=\:\frac{−\mathrm{1}\pm{i}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}}{\mathrm{2}} \\ $$$$\: \\ $$$$\boldsymbol{{Solution}}\:\boldsymbol{{by}}:\:{Kevin} \\ $$

Commented by mathdanisur last updated on 07/Aug/21

Thankyou Ser

$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com