x-1-x-2-05-x-

Question Number 204480 by EJJDJX last updated on 18/Feb/24

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

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

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

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