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sashaariefiev
@sashaariefiev
August 2022
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Помогите срочно надо!!!!!!!!!!!!!
Составить биквадратное уравнение, имеющее в числе своих корней числа √3 и √2.
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uekmyfhfp
X^4 - 5x^2 + 6 = 0;
x^2 = t;
t^2 - 5 t + 6 = 0;
D = 25 - 24 = 1;
t1 = 3; x^2 = 3; x = + -
√ 3;
t2 = 2; x = + -
√2.
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Answers & Comments
x^2 = t;
t^2 - 5 t + 6 = 0;
D = 25 - 24 = 1;
t1 = 3; x^2 = 3; x = + - √ 3;
t2 = 2; x = + - √2.