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Question Number 168878 by bounhome last updated on 20/Apr/22

solve: x^(2019) +x^(2020) +x^(2021) +x^(2022) =4

$${solve}:\:\:{x}^{\mathrm{2019}} +{x}^{\mathrm{2020}} +{x}^{\mathrm{2021}} +{x}^{\mathrm{2022}} =\mathrm{4} \\ $$

Commented by bagjagugum123 last updated on 20/Apr/22

x=1

$${x}=\mathrm{1} \\ $$

Commented by mathlove last updated on 20/Apr/22

x^(2019) +x^(2020) +x^(2021) +x^(2022) =1^(2019) +1^(2020) +1^(2021) +1^(2022) x=1

$$\:{x}^{\mathrm{2019}} +{x}^{\mathrm{2020}} +{x}^{\mathrm{2021}} +{x}^{\mathrm{2022}} =\mathrm{1}^{\mathrm{2019}} +\mathrm{1}^{\mathrm{2020}} +\mathrm{1}^{\mathrm{2021}} +\mathrm{1}^{\mathrm{2022}} \\ $$$${x}=\mathrm{1} \\ $$

Commented by MJS_new last updated on 20/Apr/22

x=1 x≈−1.0031931511975 and 2020 complex solutions

$${x}=\mathrm{1} \\ $$$${x}\approx−\mathrm{1}.\mathrm{0031931511975} \\ $$$$\mathrm{and}\:\mathrm{2020}\:\mathrm{complex}\:\mathrm{solutions} \\ $$

Answered by Rasheed.Sindhi last updated on 20/Apr/22

solve: x^(2019) +x^(2020) +x^(2021) +x^(2022) =4 (x^(2019) −1)+(x^(2020) −1) +(x^(2021) −1)+(x^(2022) −1)=0 (x−1)(...) +(x−1)(...) +(x−1)(...) +(x−1)(...) =0 (x−1)(...)=0 x−1=0 x=1

$${solve}:\:\:{x}^{\mathrm{2019}} +{x}^{\mathrm{2020}} +{x}^{\mathrm{2021}} +{x}^{\mathrm{2022}} =\mathrm{4} \\ $$$$\:\left({x}^{\mathrm{2019}} −\mathrm{1}\right)+\left({x}^{\mathrm{2020}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({x}^{\mathrm{2021}} −\mathrm{1}\right)+\left({x}^{\mathrm{2022}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$ \\ $$$$\left({x}−\mathrm{1}\right)\left(...\right) \\ $$$$\:\:\:\:\:\:\:+\left({x}−\mathrm{1}\right)\left(...\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({x}−\mathrm{1}\right)\left(...\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({x}−\mathrm{1}\right)\left(...\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0} \\ $$$$ \\ $$$$\left({x}−\mathrm{1}\right)\left(...\right)=\mathrm{0} \\ $$$${x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{1} \\ $$

Commented by mathlove last updated on 20/Apr/22

nice

$${nice} \\ $$

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