高専の数学解説

第4章 4.12

3x2+6x−7=03x^2 + 6x - 7 = 0

x=−3±9+213=−3±303x = \frac{-3 \pm \sqrt{9+21}}{3} = \frac{-3 \pm \sqrt{30}}{3}

ここで α=−3+303\alpha = \dfrac{-3+\sqrt{30}}{3}、β=−3−303\beta = \dfrac{-3-\sqrt{30}}{3} とおくと与式は

(x−α)(x−β)=0(x-\alpha)(x-\beta) = 0 と書けるので x2−(α+β)x+αβ=0x^2 - (\alpha+\beta)x + \alpha\beta = 0 となり

αβ=9−309=−73α+β=−63=−2となる\alpha\beta = \frac{9-30}{9} = -\frac{7}{3} \qquad \alpha+\beta = \frac{-6}{3} = -2 \qquad \text{となる}

(1)

α2β+αβ2=αβ(α+β)=−73⋅(−2)=143\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha+\beta) = -\frac{7}{3}\cdot(-2) = \frac{14}{3}

(2)

α3+β3=(α+β)(α2−αβ+β2)=(α+β){(α+β)2−3αβ}=(−2){(−2)2−3(−73)}=−2⋅(4+7)=−22\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2) = (\alpha+\beta)\{(\alpha+\beta)^2-3\alpha\beta\} = (-2)\left\{(-2)^2-3\left(-\frac{7}{3}\right)\right\} = -2\cdot(4+7) = -22

(3)

βα+αβ=β2+α2αβ=(α+β)2−2αβαβ=(−2)2−2(−73)−73=−267\frac{\beta}{\alpha}+\frac{\alpha}{\beta} = \frac{\beta^2+\alpha^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} = \frac{(-2)^2-2\left(-\frac{7}{3}\right)}{-\frac{7}{3}} = -\frac{26}{7}

解説: r31bn1z