数学 - Python - 代数学 - 1次方程式, 2次方程式 - 2次方程式 - 解、因数分解

学習環境

代数への出発 (新装版数学入門シリーズ) (松坂和夫(著)、岩波書店)の第4章(1次方程式, 2次方程式 )、2(2次方程式)の問9の解答を求めてみる。

1. $\begin{array}{l} (x + 4) (x + 5) = 0 \\ x = - 5, - 4 \end{array}$
2. $\begin{array}{l} x^{2} - 4 x - 12 = 0 \\ (x - 6) (x + 2) = 0 \\ x = - 2, 6 \end{array}$
3. $\begin{array}{l} x^{2} + 3 x = 0 \\ x (x + 3) = 0 \\ x = - 3, 0 \end{array}$
4. $\begin{array}{l} (2 x + 15) (2 x - 15) = 0 \\ x = \pm \frac{15}{2} \end{array}$
5. $\begin{array}{l} 4 x^{2} - 12 x + 9 = 0 \\ {(2 x - 3)}^{2} = 0 \\ x = \frac{3}{2} \end{array}$
6. $\begin{array}{l} (2 x - 5) (7 x + 2) = 0 \\ x = - \frac{2}{7}, \frac{5}{2} \end{array}$
7. $\begin{array}{l} x^{2} - 9 x + 20 = 3 x^{2} - 9 x + 6 - 4 \\ 2 x^{2} - 18 = 0 \\ x^{2} - 9 = 0 \\ x = \pm 3 \end{array}$
8. $\begin{array}{l} (x - 11 a) (x + 4 a) = 0 \\ x = - 4 a, 12 a \end{array}$
9. $\begin{array}{l} {(x + 2 a)}^{2} = 0 \\ x = - 2 a \end{array}$
10. $\begin{array}{l} x^{2} - 2 b x - (a + b) (a - b) = 0 \\ (x - (a + b)) (x + (a - b)) = 0 \\ x = \pm a - b \end{array}$

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import symbols, solveset, Rational print('9.') class MyTestCase(TestCase): def test(self): x, a, b = symbols('x, a, b', real=True) eqs = [x ** 2 + 9 * x + 20, x ** 2 - 4 * (x + 3), -x ** 2 - 3 * x, 4 * x ** 2 - 25, 4 * x ** 2 - 12 * x + 9, 14 * x ** 2 - 31 * x - 10, (x - 4) * (x - 5) - (3 * (x - 1) * (x - 2) - 4), x ** 2 - 7 * a * x - 44 * a * 2, x ** 2 + 4 * a * x + 4 * a ** 2, x ** 2 - 2 * b * x - (a ** 2 - b ** 2)] egg = [{-5, -4}, {-2, 6}, {-3, 0}, {-Rational(15, 2), Rational(15, 2)}, {Rational(3, 2)}, {-Rational(2, 7), Rational(5, 2)}, {-3, 3}, {-4 * a, 12 * a}, {-2 * a}, {a - b, -a - b}] for eq, s in zip(eqs, egg): self.assertTrue(solveset(eq, x), s) if __name__ == "__main__": main()

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2020年2月8日土曜日

数学 - Python - 代数学 - 1次方程式, 2次方程式 - 2次方程式 - 解、因数分解

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