x-1-x-2-be-the-roots-of-the-equation-x-2-x-m-0-amp-x-1-5-x-2-5-2021-Find-the-sum-of-the-possible-values-of-m-

Question Number 161066 by blackmamba last updated on 11/Dec/21

x_{1}, x_{2} b e t h e r o o t s o f t h e e q u a t i o n

x^{2} + x + m = 0 & x_{1}^{5} + x_{2}^{5} = 2021 .

F i n d t h e s u m o f t h e p o s s i b l e v a l u e s

o f m .

Answered by cortano last updated on 11/Dec/21

\Rightarrow x^{2} + x + m = 0 \to {\begin{cases} x_{1} + x_{2} = - 1 \\ x_{1} x_{2} = m \end{cases}

\Rightarrow x_{1}^{2} + x_{2}^{2} = 1 - 2 m

\Rightarrow x_{1}^{3} + x_{2}^{3} = 3 m - 1

\Rightarrow (x_{1}^{3} + x_{2}^{3}) (x_{1}^{2} + x_{2}^{2}) = (3 m - 1) (1 - 2 m)

\Rightarrow x_{1}^{5} + x_{2}^{5} + {(x_{1} x_{2})}^{2} (x_{1} + x_{2}) = - 6 m^{2} + 5 m - 1

\Rightarrow 2021 - m^{2} + 6 m^{2} - 5 m + 1 = 0

\Rightarrow 5 m^{2} - 5 m + 2022 = 0

\Rightarrow \underset{i = 1}{\sum^{2}} m_{i} = 1

Answered by TheSupreme last updated on 11/Dec/21

l e t, s x_{1} = a x_{2} = b

a b = m

a + b = 1

{(a + b)}^{5} = - 1 = a^{5} + b^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 {a b}^{4} + b^{5}

- 1 = 2021 + 5 a^{3} m - 10 m^{2} + 5 {m b}^{3}

- 2022 = 5 m (a^{3} + b^{3}) - 10 m^{2}

(a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2}) = - ({(a + b)}^{2} - 3 a b) = - (1 - 3 m) = 3 m - 1

- 2022 = 5 m (3 m - 1) - 10 m^{2}

- 2022 = 5 m^{2} - 5 m

n o s o l u t i o n s Δ < 0 .

s u m a l l p o s s i b l e v a l y e = 0

Commented by cortano last updated on 11/Dec/21

f a l s e

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