高専の数学解説

第4章 4.26

3x2−2x−4=0x2−23x−43=03x^2 - 2x - 4 = 0 \qquad x^2 - \frac{2}{3}x - \frac{4}{3} = 0

解が α、β なので与式は (x−α)(x−β)=0(x-\alpha)(x-\beta) = 0 のかたちとなるので x2−(α+β)x+αβ=0x^2 - (\alpha+\beta)x + \alpha\beta = 0

これから α+β=23\alpha + \beta = \dfrac{2}{3}、αβ=−43\alpha\beta = -\dfrac{4}{3}、α2+β2=(α+β)2−2αβ=289\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \dfrac{28}{9} となる

(1)

1α+1β=α+βαβ=23−43=−12\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{\dfrac{2}{3}}{-\dfrac{4}{3}} = -\frac{1}{2}

(2)

1(α+1)+1(β+1)=α+β+2(α+1)(β+1)=23+2αβ+(α+β)+1=83−43+23+1=83−4+2+33=8\frac{1}{(\alpha+1)} + \frac{1}{(\beta+1)} = \frac{\alpha+\beta+2}{(\alpha+1)(\beta+1)} = \frac{\dfrac{2}{3}+2}{\alpha\beta+(\alpha+\beta)+1} = \frac{\dfrac{8}{3}}{-\dfrac{4}{3}+\dfrac{2}{3}+1} = \frac{\dfrac{8}{3}}{\dfrac{-4+2+3}{3}} = 8

(3)

βα+αβ=β2+α2αβ=289−43=−73\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^2+\alpha^2}{\alpha\beta} = \frac{\dfrac{28}{9}}{-\dfrac{4}{3}} = -\frac{7}{3}

(4)

(α−β)2=α2+β2−2αβ=289−2⋅(−43)=28+249=529(\alpha-\beta)^2 = \alpha^2+\beta^2-2\alpha\beta = \frac{28}{9} - 2\cdot\left(-\frac{4}{3}\right) = \frac{28+24}{9} = \frac{52}{9}

(5)

α3+β3=(α+β)(α2−αβ+β2)=23⋅(289+43)=23⋅409=8027\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2) = \frac{2}{3}\cdot\left(\frac{28}{9}+\frac{4}{3}\right) = \frac{2}{3}\cdot\frac{40}{9} = \frac{80}{27}

(6)

(∣α∣−∣β∣)2=α2−2∣α∣∣β∣+β2=289−2⋅∣−43∣=289−83=49∣α∣−∣β∣=±23(|\alpha|-|\beta|)^2 = \alpha^2 - 2|\alpha||\beta| + \beta^2 = \frac{28}{9} - 2\cdot\left|-\frac{4}{3}\right| = \frac{28}{9} - \frac{8}{3} = \frac{4}{9} \qquad |\alpha|-|\beta| = \pm\frac{2}{3}

解説: r31bn1z