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dashasadovniche
@dashasadovniche
July 2022
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Не решая уравнения: x^2+4x-2=0, найдите значения выражения: a) x1^2+x2^2=?
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emerald0101
Verified answer
По т. Виетта х1+х2=-4, х1*х2=-2.
Возведем в квадрат (х1+х2)^2=(-4)^2 → x1^2+x2^2 +2*x1*x2=16 →
x1^2+x2^2 +2*(-2)=16 → x1^2+x2^2 =20
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Verified answer
По т. Виетта х1+х2=-4, х1*х2=-2.Возведем в квадрат (х1+х2)^2=(-4)^2 → x1^2+x2^2 +2*x1*x2=16 →
x1^2+x2^2 +2*(-2)=16 → x1^2+x2^2 =20