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Ответ

Ответ дан
papagenius

Ответ:

Объяснение:

[left{begin{gathered}{x_1}+{x_2}=-40hfill\{x_1}cdot{x_2}=-1600hfill\end{gathered}right.Leftrightarrowleft{begin{gathered}{x_1}=-40-{x_2}hfill\{x_1}cdot{x_2}=-1600hfill\end{gathered}right.Leftrightarrow]

[left{begin{gathered}{x_1}=-40-{x_2}hfill\(-40-{x_2})cdot{x_2}=-1600hfill\end{gathered}right.Leftrightarrowleft{begin{gathered}{x_1}=-40-{x_2}hfill\-40{x_2}-x_2^2+1600=0hfill\end{gathered}right.]

Решим квадратное уравнение:

[begin{gathered}-40{x_2}-x_2^2+1600=0hfill\-40x-{x^2}+1600=0;;;;|cdot (-1)hfill\{x^2}+40x-1600=0hfill\D={b^2}-4ac={40^2}+4cdot 1cdot1600=1600+6400=8000hfill\end{gathered}]

[begin{gathered}{x_{1;2}}=frac{{-bpmsqrt D}}{{2a}}=frac{{-40pmsqrt {8000}}}{{2cdot 1}}=frac{{-40pm40sqrt 5}}{2}hfill\{x_1}=frac{{-40+40sqrt 5}}{2}=frac{{2(-20+20sqrt 5)}}{2}=-20+20sqrt 5hfill\{x_2}=frac{{-40-40sqrt 5}}{2}=frac{{2(-20-20sqrt 5)}}{2}=-20-20sqrt 5hfill\end{gathered}]

Итак, мы нашли x₂.

{x_2}=-20+20sqrt 5

или

{x_2}=-20-20sqrt 5

Найдем x₁

{x_1}=-40-{x_2}hfill\

При {x_2}=-20+20sqrt 5  

[begin{gathered}{x_1}=-40-(-20+20sqrt 5)=-20-20sqrt 5hfill\end{gathered}]

При {x_2}=-20-20sqrt 5

[{x_1}=-40-(-20-20sqrt 5)=-20+20sqrt 5]

Ответ: [{x_1}=-20-20sqrt 5;;;;{x_2}=-20+20sqrt 5]   или [{x_1}=-20+20sqrt 5;{x_2}=-20-20sqrt 5]

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