3.1 一元方程#

from sympy import symbols, solve, Eq

x = symbols('x')

# 直接給運算式（預設 = 0）
print(solve(x**2 - 4, x))          # [-2, 2]

# 用 Eq 表示等式：2x + 3 = 7
print(solve(Eq(2*x + 3, 7), x))    # [2]

# 一次方程
print(solve(3*x - 5, x))           # [5/3]

from sympy import symbols, solve, sqrt, Rational, discriminant

a, b, c, x = symbols('a b c x')

# 1. 一般求根公式
roots_formula = solve(a*x**2 + b*x + c, x)
print(roots_formula)
# [(-b + sqrt(-4*a*c + b**2))/(2*a), (-b - sqrt(-4*a*c + b**2))/(2*a)]

# 2. 判別式
D = b**2 - 4*a*c
print(discriminant(a*x**2 + b*x + c, x))  # b**2 - 4*a*c，一致！

# 3. 具體例子 x^2 - 5x + 6 = 0
print(solve(x**2 - 5*x + 6, x))   # [2, 3]

# 4. 沒有實根的例子：x^2 + x + 1 = 0
print(solve(x**2 + x + 1, x))
# [-1/2 - sqrt(3)*I/2, -1/2 + sqrt(3)*I/2]  → 複根，精確表示

# 5. 含無理根的例子：x^2 - 2 = 0
print(solve(x**2 - 2, x))         # [-sqrt(2), sqrt(2)]

from sympy import symbols, solve, roots, Poly, nsimplify, N

x = symbols('x')

# 三次方程：根其實是 4, -2±sqrt(3)
sol = solve(x**3 - 15*x - 4, x)
print(sol)
# [4, -sqrt(3) - 2, sqrt(3) - 2]   ← 這裡 SymPy 已化簡得不錯

# 換一個更狡猾的例子：x^3 - 3x - 1 = 0（三個實根，但全是醜的無理數）
sol2 = solve(x**3 - 3*x - 1, x)
print(sol2)
# 2*cos(...) 形式或冗長根式，視版本而定
for r in sol2:
    print(r, '≈', N(r, 6))

from sympy import symbols, roots, Poly, solve

x = symbols('x')

# roots() 回傳 {根: 重數}
print(roots((x - 1)**2 * (x + 3), x))
# {1: 2, -3: 1}   ← 1 是二重根

# Poly 物件：明確宣告變數與域
p = Poly(x**4 - 5*x**2 + 4, x)
print(p)                # Poly(x**4 - 5*x**2 + 4, x, domain='ZZ')
print(p.degree())       # 4
print(p.all_roots())    # [-2, -1, 1, 2]（按代數數排好）
print(p.factor_list())  # 因式分解

# 高次方程：4 次還有公式，5 次以上一般沒有
print(roots(x**5 - x - 1, x))   # {} ← 根式解不存在，只能靠數值方法

from sympy import symbols, solve, Abs, sqrt, checksol

x = symbols('x')

# 無理方程：sqrt(x+3) = x - 3
# 手算：兩邊平方 → x + 3 = x^2 - 6x + 9 → x^2 - 7x + 6 = 0 → x = 1 或 6
# 但平方會產生「增根」：x=1 代回原式 sqrt(4)=2 ≠ -2，捨去！
print(solve(sqrt(x + 3) - (x - 3), x))   # [6] ← SymPy 自動檢查，只留 6

# 保險起見可手動驗根
for cand in [1, 6]:
    print(cand, checksol(sqrt(x + 3) - (x - 3), x, cand))  # None / 0

# 絕對值方程：|2x - 1| = 5
# 手算：2x - 1 = 5 或 2x - 1 = -5 → x = 3 或 x = -2
print(solve(Abs(2*x - 1) - 5, x))        # [-2, 3]

# 含兩個絕對值：|x| + |x - 1| = 3（分段討論手算很繁瑣）
print(solve(Abs(x) + Abs(x - 1) - 3, x)) # [-1, 2]

from sympy import symbols, solve, discriminant, N, Rational

x, c = symbols('x c')
a, b = 1, 2

print(f"{'c':>6} | {'Δ':>8} | 根")
print("-" * 50)
for cv in [Rational(3), Rational(1), Rational(0), Rational(-1), Rational(-2)]:
    poly = a*x**2 + b*x + cv - cv*0 + (cv - cv)  # 展開寫法避免混淆
    poly = a*x**2 + b*x + cv
    D = discriminant(poly, x)
    rts = solve(poly, x)
    print(f"{str(cv):>6} | {str(D):>8} | {rts}")

# 結論：
# c = 1    → Δ = 0     → 重根 [-1]
# c = 0    → Δ = 4 > 0 → 兩實根 [0, -2]
# c = -1   → Δ = 8 > 0 → 兩實根（無理）
# c = 3    → Δ = -8 < 0→ 兩複根

# 實驗延伸：讓 c 保持符號，寫出「有實根」的條件
from sympy import simplify
D_general = b**2 - 4*a*c
# 對 a=1, b=2：Δ = 4 - 4c ≥ 0 ⟺ c ≤ 1
print(solve(D_general.subs({a: 1, b: 2}) >= 0, c))  # c ≤ 1（見 3.3 節不等式）