数学 – 数学の威力を発揮する - 方程式 – 高次方程式(連立2次方程式, いろいろな解き方)

学習環境

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32.を解いてみる。

問32.

$\begin{array}{l} y = x - 1 \\ x^{2} + {(x - 1)}^{2} = 25 \\ 2 x^{2} - 2 x - 24 = 0 \\ x^{2} - x - 12 = 0 \\ (x - 4) (x + 3) = 0 \\ x = 4, y = 3; \\ x = - 3, y = - 4 \end{array}$
$\begin{array}{l} x^{2} + 3 x^{2} = 48 \\ x^{2} = 12 \\ x = \pm 2 \sqrt{3}, y = \pm 6 \end{array}$
$\begin{array}{l} {(2 x - 1)}^{2} - x^{2} = 5 \\ 3 x^{2} - 4 x - 4 = 0 \\ (x - 2) (3 x + 2) = 0 \\ x = 2, y = 3; \\ x = - \frac{2}{3}, y = - \frac{7}{3} \end{array}$
$\begin{array}{l} t^{2} - 4 t + 2 = 0 \\ t = 2 \pm \sqrt{2} \\ x = 2 \pm \sqrt{2}, y = 2 \mp \sqrt{2} \end{array}$
$\begin{array}{l} {(x + y)}^{2} - x y = 21 \\ 25 - x y = 21 \\ x y = 4 \\ t^{2} - 5 t + 4 = 0 \\ (t - 1) (t - 4) = 0 \\ x = 1, y = 4; \\ x = 4, y = 1 \end{array}$
$\begin{array}{l} (x + y) (4 x - 3 y) = 0 \\ y = - x \\ 3 x - 4 x = 5 \\ x = - 5, y = 5; \\ y = \frac{4 x}{3} \\ 3 x + \frac{16 x}{3} = 5 \\ 25 x = 15 \\ x = \frac{3}{5}, y = \frac{4}{5} \end{array}$

Kamimura's blog