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 204480 by EJJDJX last updated on 18/Feb/24

x +  (1/x) = 2.05  x = ?

$${x}\:+\:\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}.\mathrm{05} \\ $$$${x}\:=\:? \\ $$

Commented by TonyCWX08 last updated on 19/Feb/24

  x^2 +1= 2.05x  x^2 −2.05x= = −1  x^2 −2.05x+(((2.05)/2))^2 = −1+(((2.05)/2))^2   (x−((2.05)/2))^2  = 0.050625  x−1.025 = ±(√(0.050625))  x = 1.025±0.225  x_1 =1.25 x_2 =0.8

$$ \\ $$$${x}^{\mathrm{2}} +\mathrm{1}=\:\mathrm{2}.\mathrm{05}{x} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}.\mathrm{05}{x}=\:=\:−\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}.\mathrm{05}{x}+\left(\frac{\mathrm{2}.\mathrm{05}}{\mathrm{2}}\right)^{\mathrm{2}} =\:−\mathrm{1}+\left(\frac{\mathrm{2}.\mathrm{05}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({x}−\frac{\mathrm{2}.\mathrm{05}}{\mathrm{2}}\right)^{\mathrm{2}} \:=\:\mathrm{0}.\mathrm{050625} \\ $$$${x}−\mathrm{1}.\mathrm{025}\:=\:\pm\sqrt{\mathrm{0}.\mathrm{050625}} \\ $$$${x}\:=\:\mathrm{1}.\mathrm{025}\pm\mathrm{0}.\mathrm{225} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}.\mathrm{25}\:{x}_{\mathrm{2}} =\mathrm{0}.\mathrm{8} \\ $$

Answered by MM42 last updated on 18/Feb/24

((x^2 +1)/x)=((41)/(20))⇒20x^2 −41x+20=0  x=1.25 & x=0.8

$$\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}=\frac{\mathrm{41}}{\mathrm{20}}\Rightarrow\mathrm{20}{x}^{\mathrm{2}} −\mathrm{41}{x}+\mathrm{20}=\mathrm{0} \\ $$$${x}=\mathrm{1}.\mathrm{25}\:\&\:{x}=\mathrm{0}.\mathrm{8} \\ $$$$ \\ $$

Answered by BaliramKumar last updated on 19/Feb/24

x + (1/x) = 2.05              ...............       (i)  x − (1/x) = ±(√(2.05^2  − 4))  = ±(√(4.2025−4))   x − (1/x) = ±(√(0.2025)) = ± 0.45             x − (1/x) =  0.45                   ...............(ii)  x − (1/x) =  −0.45           .......................(iii)  (i)+(ii)  2x = 2.5        ⇒  x = 1.25  (i)+(iii)  2x = 1.6        ⇒ x = 0.8

$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}.\mathrm{05}\:\:\:\:\:\:\:\:\:\:\:\:\:\:...............\:\:\:\:\:\:\:\left({i}\right) \\ $$$${x}\:−\:\frac{\mathrm{1}}{{x}}\:=\:\pm\sqrt{\mathrm{2}.\mathrm{05}^{\mathrm{2}} \:−\:\mathrm{4}}\:\:=\:\pm\sqrt{\mathrm{4}.\mathrm{2025}−\mathrm{4}}\: \\ $$$${x}\:−\:\frac{\mathrm{1}}{{x}}\:=\:\pm\sqrt{\mathrm{0}.\mathrm{2025}}\:=\:\pm\:\mathrm{0}.\mathrm{45}\:\:\:\:\:\:\:\:\:\:\: \\ $$$${x}\:−\:\frac{\mathrm{1}}{{x}}\:=\:\:\mathrm{0}.\mathrm{45}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...............\left({ii}\right) \\ $$$${x}\:−\:\frac{\mathrm{1}}{{x}}\:=\:\:−\mathrm{0}.\mathrm{45}\:\:\:\:\:\:\:\:\:\:\:.......................\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right) \\ $$$$\mathrm{2}{x}\:=\:\mathrm{2}.\mathrm{5}\:\:\:\:\:\:\:\:\Rightarrow\:\:{x}\:=\:\mathrm{1}.\mathrm{25} \\ $$$$\left({i}\right)+\left({iii}\right) \\ $$$$\mathrm{2}{x}\:=\:\mathrm{1}.\mathrm{6}\:\:\:\:\:\:\:\:\Rightarrow\:{x}\:=\:\mathrm{0}.\mathrm{8} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com