Question
If x = \dfrac{\sqrt{2019} +1 }{2}
Find the value of
(4x^3 - 2022x-2019)^{2022}
If x = \dfrac{\sqrt{2019} +1 }{2}
Find the value of
(4x^3 - 2022x-2019)^{2022}
x = \dfrac{\sqrt{2019} +1 }{2}
x^2 = \dfrac{1}{4} (2020+2\sqrt{2019} ) = \dfrac{1}{2}(1010+\sqrt{2019} )
4x^3 - 2022x-2019
=4x\cdot \dfrac{1}{2}(1010+\sqrt{2019} )- 2022x-2019
=2x(1010+\sqrt{2019} )- 2022x-2019
=2020x+2x\sqrt{2019} - 2022x-2019
=2x\sqrt{2019}-2x-2019
=2x(\sqrt{2019}-1 )-2019
=2\cdot \dfrac{\sqrt{2019} +1 }{2} \cdot(\sqrt{2019}-1 )-2019
=2019-1-2019
=-1
Therefore
(4x^3 - 2022x-2019)^{2022}
=(-1)^{2022} =1