Question

произведение корней уравнения...(x^2+x+1)^2-3x^2-3=3

Answer 1

Подсказка:

 

Раскрываем все скобки, представляем в виде многочлена и правниваем к нулю.

 

Если задание правильное (проверьте!), то это уравнение равносильно уравнению

(х - 1)(x^3 + 3x^2 + 3x + 5) = 0.

 

Если задание правильное, продолжим решение в чате.

Answer 2

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The user's message is a request for a detailed answer to the product of the roots of the equation $(x^2+x+1)^2-3x^2-3=3$. I will use the web search results from my predefined internal tool to provide a possible solution.

The equation can be rewritten as $(x^2+x+1)^2=3x^2+6$, which is equivalent to $(x^2+x+1-3x)(x^2+x+1+3x)=6$. To find the roots of this equation, we need to solve the two quadratic factors separately.

The first factor is $x^2+x+1-3x=x^2-2x+1$, which has a discriminant of $D=(-2)^2-4\cdot 1\cdot 1=0$. Therefore, it has a single real root of $x_1=\frac{-(-2)}{2\cdot 1}=1$.

The second factor is $x^2+x+1+3x=x^2+4x+1$, which has a discriminant of $D=4^2-4\cdot 1\cdot 1=12$. Therefore, it has two distinct real roots of $x_2=\frac{-4-\sqrt{12}}{2\cdot 1}=-2-\sqrt{3}$ and $x_3=\frac{-4+\sqrt{12}}{2\cdot 1}=-2+\sqrt{3}$.

The product of the roots of the equation is then $x_1\cdot x_2\cdot x_3=1\cdot (-2-\sqrt{3})\cdot (-2+\sqrt{3})=1\cdot (4-3)=-1$.

Therefore, the answer is -1. You can also check other sources for more explanations and examples , , , , or .

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