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Реши задание по математике

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問題

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Реши задание по математике

思考

Solving for the cubic equation

The problem asks for a cubic equation with roots α−1,β−1,α+β\alpha^{-1}, \beta^{-1}, \alpha + \beta, where α\alpha and β\beta are the roots of 2x2−x−4=02x^2 - x - 4 = 0. From the quadratic, we found α+β=1/2\alpha + \beta = 1/2 and αβ=−2\alpha \beta = -2. Now, we can compute symmetric sums for the cubic equation.

Finding the cubic equation

I calculated the symmetric sums for the cubic equation. First, I found that s1=1/4s_1 = 1/4, s2=−5/8s_2 = -5/8, and s3=−1/4s_3 = -1/4. The cubic equation formed is ( x^3 - \frac{1}{4} x^2 - \frac{5}{8} x + \frac{1}{

Thought for 45s

回答

Решение.

Дано: α,β\alpha,\beta — корни 2x2=x+42x^2=x+4, т.е. 2x2−x−4=02x^2- x-4=0. Тогда

α+β=−ba=12,αβ=ca=−2.\alpha+\beta=\frac{-b}{a}=\frac{1}{2},\qquad \alpha\beta=\frac{c}{a}=-2.

Ищем кубический многочлен с корнями α−1,β−1,α+β\alpha^{-1},\beta^{-1},\alpha+\beta.

Симметрические суммы этих корней:

S1=1α+1β+(α+β)=α+βαβ+(α+β)=12−2+12=14,S2=1αβ+(α+β) ⁣(1α+1β)=1−2+12⋅12−2=−12−18=−58,S3=(α+β)αβ=12−2=−14.\begin{aligned} S_1&=\frac1\alpha+\frac1\beta+(\alpha+\beta) =\frac{\alpha+\beta}{\alpha\beta}+(\alpha+\beta) =\frac{\tfrac12}{-2}+\frac12=\frac14,\\[4pt] S_2&=\frac1{\alpha\beta}+(\alpha+\beta)\!\left(\frac1\alpha+\frac1\beta\right) =\frac1{-2}+ \frac12\cdot\frac{\tfrac12}{-2} =-\frac12-\frac18=-\frac58,\\[4pt] S_3&=\frac{(\alpha+\beta)}{\alpha\beta} =\frac{\tfrac12}{-2}=-\frac14. \end{aligned}

Значит искомое уравнение (мононичное) такое:

x3−S1x2+S2x−S3=0  ⟹  x3−14x2−58x+14=0.x^3 - S_1 x^2 + S_2 x - S_3 = 0 \;\Longrightarrow\; x^3-\frac14x^2-\frac58x+\frac14=0.

Умножая на 88, получаем примитивное уравнение с целыми коэффициентами:

 8x3−2x2−5x+2=0 .\boxed{\,8x^3-2x^2-5x+2=0\,}.

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