数学 – 数学の威力を発揮する - 方程式 – 高次方程式(3元以上の連立2次方程式)

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数学読本〈1〉数・方程式/方程式/不等式 (松坂和夫(著)、岩波書店)の第3章(数学の威力を発揮する - 方程式)、3.3(高次方程式)、連立2次方程式の問38.を解いてみる。

問38.

$\begin{array}{l} y = - 1 - 2 x \\ x + 2 + 4 x + z = - 4 \\ z = - 5 x - 6 \\ x^{2} + 1 + 4 x^{2} + 4 x + 25 x^{2} + 36 + 60 x = 29 \\ 30 x^{2} + 64 x + 8 = 0 \\ 15 x^{2} + 32 x + 4 = 0 \\ x = \frac{- 16 \pm \sqrt{196}}{15} \\ = \frac{- 16 \pm 14}{15} \\ x = - \frac{2}{15}, y = - \frac{11}{15}, z = - \frac{16}{3} \\ x = - 2, y = 3, z = 4 \end{array}$
$\begin{array}{l} y = \frac{420}{x} \\ z = 70 - \frac{420}{x} - x \\ x^{2} + {(\frac{420}{x})}^{2} = {(70 - \frac{420}{x} - x)}^{2} \\ 70^{2} - \frac{2 \cdot 70 \cdot 420}{x} + 2 \cdot 420 - 2 \cdot 70 x = 0 \\ 70 x - 2 \cdot 420 + 2 \cdot 6 x - 2 x^{2} = 0 \\ x^{2} - 41 x + 420 = 0 \\ (x - 20) (x - 21) = 0 \\ x = 20, y = 21, z = 29 \\ x = 21, y = 20, z = 29 \end{array}$
$\begin{array}{l} {(x + y + z)}^{2} = 16 \\ x + y + z = \pm 4 \\ x = \pm \frac{3}{2}, y = \mp \frac{5}{2}, z = \pm 5 \end{array}$
$\begin{array}{l} X = y z, Y = z x, Z = x y \\ 2 X + Y = 0 \\ Y - 3 Z = 0 \\ Z + 4 X = 10 \\ Y = - 2 X \\ Z = 10 - 4 X \\ - 2 X - 30 + 12 X = 0 \\ 10 X = 30 \\ X = 3 \\ Y = - 6 \\ Z = - 2 \\ y z = 3, z x = - 6, x y = - 2 \\ y = - \frac{2}{x} \\ z = - \frac{6}{x} \\ \frac{12}{x^{2}} = 3 \\ x = \pm 2, y = \mp 1, z = \mp 3 \end{array}$

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