高専の数学解説

第4章 4.27

(1)

2x2−3x−4=02x^2 - 3x - 4 = 0

x=3±9+324=3±414x = \frac{3 \pm \sqrt{9+32}}{4} = \frac{3 \pm \sqrt{41}}{4}

2α−1=3+412−1=1+4122β−1=3−412−1=1−4122\alpha - 1 = \frac{3+\sqrt{41}}{2} - 1 = \frac{1+\sqrt{41}}{2} \qquad 2\beta - 1 = \frac{3-\sqrt{41}}{2} - 1 = \frac{1-\sqrt{41}}{2}

この２つが解なので

(2x−1−41)(2x−1+41)=(2x−1)2−41=4x2−4x−40=0x2−x−10=0(2x-1-\sqrt{41})(2x-1+\sqrt{41}) = (2x-1)^2 - 41 = 4x^2-4x-40=0 \qquad x^2-x-10=0

(2)

βα−1=3−4143+414−1=3−413+41−4=3−41−1+41=−3−411−41\frac{\beta}{\alpha-1} = \frac{\frac{3-\sqrt{41}}{4}}{\frac{3+\sqrt{41}}{4}-1} = \frac{3-\sqrt{41}}{3+\sqrt{41}-4} = \frac{3-\sqrt{41}}{-1+\sqrt{41}} = -\frac{3-\sqrt{41}}{1-\sqrt{41}}

αβ−1=3+4143−414−1=3+413−41−4=3+41−1−41=−3+411+41\frac{\alpha}{\beta-1} = \frac{\frac{3+\sqrt{41}}{4}}{\frac{3-\sqrt{41}}{4}-1} = \frac{3+\sqrt{41}}{3-\sqrt{41}-4} = \frac{3+\sqrt{41}}{-1-\sqrt{41}} = -\frac{3+\sqrt{41}}{1+\sqrt{41}}

この２つが解なので

(x+3−411−41)(x+3+411+41)=x2+(3−411−41+3+411+41)x+3−411−41⋅3+411+41\left(x+\frac{3-\sqrt{41}}{1-\sqrt{41}}\right)\left(x+\frac{3+\sqrt{41}}{1+\sqrt{41}}\right) = x^2 + \left(\frac{3-\sqrt{41}}{1-\sqrt{41}}+\frac{3+\sqrt{41}}{1+\sqrt{41}}\right)x + \frac{3-\sqrt{41}}{1-\sqrt{41}}\cdot\frac{3+\sqrt{41}}{1+\sqrt{41}}

=x2+(3−41)(1+41)+(3+41)(1−41)1−41x+9−411−41= x^2 + \frac{(3-\sqrt{41})(1+\sqrt{41})+(3+\sqrt{41})(1-\sqrt{41})}{1-41}x + \frac{9-41}{1-41}

=x2−3+241−41+3−241−4140x+3240=x2+1910x+45=010x2+19x+8=0= x^2 - \frac{3+2\sqrt{41}-41+3-2\sqrt{41}-41}{40}x + \frac{32}{40} = x^2 + \frac{19}{10}x + \frac{4}{5} = 0 \qquad 10x^2+19x+8=0

解説: r31bn1z