Given-x-3-3x-2-2-6x-2-2-8-0-then-x-5-41x-2-2022-

Question Number 166954 by cortano1 last updated on 03/Mar/22

Given x^{3} - {3 x}^{2} \sqrt{2} + 6 x - 2 \sqrt{2} - 8 = 0

then x^{5} - {41 x}^{2} + 2022 = ?

Answered by som(math1967) last updated on 03/Mar/22

x^{3} - 3 . x^{2} . \sqrt{2} + 3 . x . {(\sqrt{2})}^{2} - {(\sqrt{2})}^{3} = 8

\Rightarrow {(x - \sqrt{2})}^{3} = 2^{3}

\Rightarrow x = 2 + \sqrt{2}

∴ x^{5} - 41 x^{2} + 2022

= {(2 + \sqrt{2})}^{5} - 41 {(2 + \sqrt{2})}^{2} + 2022

= 32 + 5 \times 16 \times \sqrt{2} + 10 \times 8 \times 2 + 10 \times 4 \times 2 \sqrt{2}

+ 5 \times 2 \times 4 + 4 \sqrt{2} - 41 (4 + 4 \sqrt{2} + 2) + 2022

= 32 + 160 + 40 + 80 \sqrt{2} + 80 \sqrt{2} + 4 \sqrt{2}

- 246 - 164 \sqrt{2} + 2022

= 2022 - 14 = 2008