Предмет: Алгебра, автор: koliagorbatuk

знайдіть всі корені квадратного тричлена x²+3x-4


срочно поможіть дам 40б​


ВикаБач: х1=-4; х2=1; (Например, по т. Виета).

Ответы

Автор ответа: bel72777
0

Ответ:

Объяснение:

По теореме Виета:

ax²+bx+c=0

x₁+x₂=-b/a; x₁·x₂=c/a

x²+3x-4=0

x₁+x₂=-3/1=-3; 1+(-4)=-3

x₁·x₂=-4/1=-4; 1·(-4)=-4

Ответ: x₁=1; x₂=-4.

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