Metamath Proof Explorer


Theorem quad2

Description: The quadratic equation, without specifying the particular branch D to the square root. (Contributed by Mario Carneiro, 23-Apr-2015)

Ref Expression
Hypotheses quad.a ⊢ φ → A ∈ ℂ
quad.z ⊢ φ → A ≠ 0
quad.b ⊢ φ → B ∈ ℂ
quad.c ⊢ φ → C ∈ ℂ
quad.x ⊢ φ → X ∈ ℂ
quad2.d ⊢ φ → D ∈ ℂ
quad2.2 ⊢ φ → D 2 = B 2 − 4 ⁢ A ⁢ C
Assertion quad2 ⊢ φ → A ⁢ X 2 + B ⁢ X + C = 0 ↔ X = - B + D 2 ⁢ A ∨ X = - B - D 2 ⁢ A

Proof

Step Hyp Ref Expression
1 quad.a ⊢ φ → A ∈ ℂ
2 quad.z ⊢ φ → A ≠ 0
3 quad.b ⊢ φ → B ∈ ℂ
4 quad.c ⊢ φ → C ∈ ℂ
5 quad.x ⊢ φ → X ∈ ℂ
6 quad2.d ⊢ φ → D ∈ ℂ
7 quad2.2 ⊢ φ → D 2 = B 2 − 4 ⁢ A ⁢ C
8 2cn ⊢ 2 ∈ ℂ
9 mulcl ⊢ 2 ∈ ℂ ∧ A ∈ ℂ → 2 ⁢ A ∈ ℂ
10 8 1 9 sylancr ⊢ φ → 2 ⁢ A ∈ ℂ
11 10 5 mulcld ⊢ φ → 2 ⁢ A ⁢ X ∈ ℂ
12 11 3 addcld ⊢ φ → 2 ⁢ A ⁢ X + B ∈ ℂ
13 12 sqcld ⊢ φ → 2 ⁢ A ⁢ X + B 2 ∈ ℂ
14 6 sqcld ⊢ φ → D 2 ∈ ℂ
15 13 14 subeq0ad ⊢ φ → 2 ⁢ A ⁢ X + B 2 − D 2 = 0 ↔ 2 ⁢ A ⁢ X + B 2 = D 2
16 5 sqcld ⊢ φ → X 2 ∈ ℂ
17 1 16 mulcld ⊢ φ → A ⁢ X 2 ∈ ℂ
18 3 5 mulcld ⊢ φ → B ⁢ X ∈ ℂ
19 18 4 addcld ⊢ φ → B ⁢ X + C ∈ ℂ
20 17 19 addcld ⊢ φ → A ⁢ X 2 + B ⁢ X + C ∈ ℂ
21 0cnd ⊢ φ → 0 ∈ ℂ
22 4cn ⊢ 4 ∈ ℂ
23 mulcl ⊢ 4 ∈ ℂ ∧ A ∈ ℂ → 4 ⁢ A ∈ ℂ
24 22 1 23 sylancr ⊢ φ → 4 ⁢ A ∈ ℂ
25 22 a1i ⊢ φ → 4 ∈ ℂ
26 4ne0 ⊢ 4 ≠ 0
27 26 a1i ⊢ φ → 4 ≠ 0
28 25 1 27 2 mulne0d ⊢ φ → 4 ⁢ A ≠ 0
29 20 21 24 28 mulcand ⊢ φ → 4 ⁢ A ⁢ A ⁢ X 2 + B ⁢ X + C = 4 ⁢ A ⋅ 0 ↔ A ⁢ X 2 + B ⁢ X + C = 0
30 11 sqcld ⊢ φ → 2 ⁢ A ⁢ X 2 ∈ ℂ
31 11 3 mulcld ⊢ φ → 2 ⁢ A ⁢ X ⁢ B ∈ ℂ
32 mulcl ⊢ 2 ∈ ℂ ∧ 2 ⁢ A ⁢ X ⁢ B ∈ ℂ → 2 ⁢ 2 ⁢ A ⁢ X ⁢ B ∈ ℂ
33 8 31 32 sylancr ⊢ φ → 2 ⁢ 2 ⁢ A ⁢ X ⁢ B ∈ ℂ
34 1 4 mulcld ⊢ φ → A ⁢ C ∈ ℂ
35 mulcl ⊢ 4 ∈ ℂ ∧ A ⁢ C ∈ ℂ → 4 ⁢ A ⁢ C ∈ ℂ
36 22 34 35 sylancr ⊢ φ → 4 ⁢ A ⁢ C ∈ ℂ
37 30 33 36 addassd ⊢ φ → 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B + 4 ⁢ A ⁢ C = 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B + 4 ⁢ A ⁢ C
38 3 sqcld ⊢ φ → B 2 ∈ ℂ
39 30 33 addcld ⊢ φ → 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B ∈ ℂ
40 38 39 36 pnncand ⊢ φ → B 2 + 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B - B 2 − 4 ⁢ A ⁢ C = 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B + 4 ⁢ A ⁢ C
41 10 5 sqmuld ⊢ φ → 2 ⁢ A ⁢ X 2 = 2 ⁢ A 2 ⁢ X 2
42 sq2 ⊢ 2 2 = 4
43 42 a1i ⊢ φ → 2 2 = 4
44 1 sqvald ⊢ φ → A 2 = A ⁢ A
45 43 44 oveq12d ⊢ φ → 2 2 ⁢ A 2 = 4 ⁢ A ⁢ A
46 sqmul ⊢ 2 ∈ ℂ ∧ A ∈ ℂ → 2 ⁢ A 2 = 2 2 ⁢ A 2
47 8 1 46 sylancr ⊢ φ → 2 ⁢ A 2 = 2 2 ⁢ A 2
48 25 1 1 mulassd ⊢ φ → 4 ⁢ A ⁢ A = 4 ⁢ A ⁢ A
49 45 47 48 3eqtr4d ⊢ φ → 2 ⁢ A 2 = 4 ⁢ A ⁢ A
50 49 oveq1d ⊢ φ → 2 ⁢ A 2 ⁢ X 2 = 4 ⁢ A ⁢ A ⁢ X 2
51 24 1 16 mulassd ⊢ φ → 4 ⁢ A ⁢ A ⁢ X 2 = 4 ⁢ A ⁢ A ⁢ X 2
52 41 50 51 3eqtrrd ⊢ φ → 4 ⁢ A ⁢ A ⁢ X 2 = 2 ⁢ A ⁢ X 2
53 24 18 4 adddid ⊢ φ → 4 ⁢ A ⁢ B ⁢ X + C = 4 ⁢ A ⁢ B ⁢ X + 4 ⁢ A ⁢ C
54 2t2e4 ⊢ 2 ⋅ 2 = 4
55 54 oveq1i ⊢ 2 ⋅ 2 ⁢ A = 4 ⁢ A
56 8 a1i ⊢ φ → 2 ∈ ℂ
57 56 56 1 mulassd ⊢ φ → 2 ⋅ 2 ⁢ A = 2 ⁢ 2 ⁢ A
58 55 57 eqtr3id ⊢ φ → 4 ⁢ A = 2 ⁢ 2 ⁢ A
59 58 oveq1d ⊢ φ → 4 ⁢ A ⁢ B = 2 ⁢ 2 ⁢ A ⁢ B
60 56 10 3 mulassd ⊢ φ → 2 ⁢ 2 ⁢ A ⁢ B = 2 ⁢ 2 ⁢ A ⁢ B
61 59 60 eqtrd ⊢ φ → 4 ⁢ A ⁢ B = 2 ⁢ 2 ⁢ A ⁢ B
62 61 oveq1d ⊢ φ → 4 ⁢ A ⁢ B ⁢ X = 2 ⁢ 2 ⁢ A ⁢ B ⁢ X
63 10 3 mulcld ⊢ φ → 2 ⁢ A ⁢ B ∈ ℂ
64 56 63 5 mulassd ⊢ φ → 2 ⁢ 2 ⁢ A ⁢ B ⁢ X = 2 ⁢ 2 ⁢ A ⁢ B ⁢ X
65 62 64 eqtrd ⊢ φ → 4 ⁢ A ⁢ B ⁢ X = 2 ⁢ 2 ⁢ A ⁢ B ⁢ X
66 24 3 5 mulassd ⊢ φ → 4 ⁢ A ⁢ B ⁢ X = 4 ⁢ A ⁢ B ⁢ X
67 10 3 5 mul32d ⊢ φ → 2 ⁢ A ⁢ B ⁢ X = 2 ⁢ A ⁢ X ⁢ B
68 67 oveq2d ⊢ φ → 2 ⁢ 2 ⁢ A ⁢ B ⁢ X = 2 ⁢ 2 ⁢ A ⁢ X ⁢ B
69 65 66 68 3eqtr3d ⊢ φ → 4 ⁢ A ⁢ B ⁢ X = 2 ⁢ 2 ⁢ A ⁢ X ⁢ B
70 25 1 4 mulassd ⊢ φ → 4 ⁢ A ⁢ C = 4 ⁢ A ⁢ C
71 69 70 oveq12d ⊢ φ → 4 ⁢ A ⁢ B ⁢ X + 4 ⁢ A ⁢ C = 2 ⁢ 2 ⁢ A ⁢ X ⁢ B + 4 ⁢ A ⁢ C
72 53 71 eqtrd ⊢ φ → 4 ⁢ A ⁢ B ⁢ X + C = 2 ⁢ 2 ⁢ A ⁢ X ⁢ B + 4 ⁢ A ⁢ C
73 52 72 oveq12d ⊢ φ → 4 ⁢ A ⁢ A ⁢ X 2 + 4 ⁢ A ⁢ B ⁢ X + C = 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B + 4 ⁢ A ⁢ C
74 37 40 73 3eqtr4rd ⊢ φ → 4 ⁢ A ⁢ A ⁢ X 2 + 4 ⁢ A ⁢ B ⁢ X + C = B 2 + 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B - B 2 − 4 ⁢ A ⁢ C
75 24 17 19 adddid ⊢ φ → 4 ⁢ A ⁢ A ⁢ X 2 + B ⁢ X + C = 4 ⁢ A ⁢ A ⁢ X 2 + 4 ⁢ A ⁢ B ⁢ X + C
76 binom2 ⊢ 2 ⁢ A ⁢ X ∈ ℂ ∧ B ∈ ℂ → 2 ⁢ A ⁢ X + B 2 = 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B + B 2
77 11 3 76 syl2anc ⊢ φ → 2 ⁢ A ⁢ X + B 2 = 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B + B 2
78 39 38 77 comraddd ⊢ φ → 2 ⁢ A ⁢ X + B 2 = B 2 + 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B
79 78 7 oveq12d ⊢ φ → 2 ⁢ A ⁢ X + B 2 − D 2 = B 2 + 2 ⁢ A ⁢ X 2 + 2 ⁢ 2 ⁢ A ⁢ X ⁢ B - B 2 − 4 ⁢ A ⁢ C
80 74 75 79 3eqtr4d ⊢ φ → 4 ⁢ A ⁢ A ⁢ X 2 + B ⁢ X + C = 2 ⁢ A ⁢ X + B 2 − D 2
81 24 mul01d ⊢ φ → 4 ⁢ A ⋅ 0 = 0
82 80 81 eqeq12d ⊢ φ → 4 ⁢ A ⁢ A ⁢ X 2 + B ⁢ X + C = 4 ⁢ A ⋅ 0 ↔ 2 ⁢ A ⁢ X + B 2 − D 2 = 0
83 29 82 bitr3d ⊢ φ → A ⁢ X 2 + B ⁢ X + C = 0 ↔ 2 ⁢ A ⁢ X + B 2 − D 2 = 0
84 11 3 subnegd ⊢ φ → 2 ⁢ A ⁢ X − − B = 2 ⁢ A ⁢ X + B
85 84 oveq1d ⊢ φ → 2 ⁢ A ⁢ X − − B 2 = 2 ⁢ A ⁢ X + B 2
86 85 eqeq1d ⊢ φ → 2 ⁢ A ⁢ X − − B 2 = D 2 ↔ 2 ⁢ A ⁢ X + B 2 = D 2
87 15 83 86 3bitr4d ⊢ φ → A ⁢ X 2 + B ⁢ X + C = 0 ↔ 2 ⁢ A ⁢ X − − B 2 = D 2
88 3 negcld ⊢ φ → − B ∈ ℂ
89 11 88 subcld ⊢ φ → 2 ⁢ A ⁢ X − − B ∈ ℂ
90 sqeqor ⊢ 2 ⁢ A ⁢ X − − B ∈ ℂ ∧ D ∈ ℂ → 2 ⁢ A ⁢ X − − B 2 = D 2 ↔ 2 ⁢ A ⁢ X − − B = D ∨ 2 ⁢ A ⁢ X − − B = − D
91 89 6 90 syl2anc ⊢ φ → 2 ⁢ A ⁢ X − − B 2 = D 2 ↔ 2 ⁢ A ⁢ X − − B = D ∨ 2 ⁢ A ⁢ X − − B = − D
92 11 88 6 subaddd ⊢ φ → 2 ⁢ A ⁢ X − − B = D ↔ - B + D = 2 ⁢ A ⁢ X
93 88 6 addcld ⊢ φ → - B + D ∈ ℂ
94 2ne0 ⊢ 2 ≠ 0
95 94 a1i ⊢ φ → 2 ≠ 0
96 56 1 95 2 mulne0d ⊢ φ → 2 ⁢ A ≠ 0
97 93 10 5 96 divmuld ⊢ φ → - B + D 2 ⁢ A = X ↔ 2 ⁢ A ⁢ X = - B + D
98 eqcom ⊢ X = - B + D 2 ⁢ A ↔ - B + D 2 ⁢ A = X
99 eqcom ⊢ - B + D = 2 ⁢ A ⁢ X ↔ 2 ⁢ A ⁢ X = - B + D
100 97 98 99 3bitr4g ⊢ φ → X = - B + D 2 ⁢ A ↔ - B + D = 2 ⁢ A ⁢ X
101 92 100 bitr4d ⊢ φ → 2 ⁢ A ⁢ X − − B = D ↔ X = - B + D 2 ⁢ A
102 88 6 negsubd ⊢ φ → - B + − D = - B - D
103 102 eqeq1d ⊢ φ → - B + − D = 2 ⁢ A ⁢ X ↔ - B - D = 2 ⁢ A ⁢ X
104 6 negcld ⊢ φ → − D ∈ ℂ
105 11 88 104 subaddd ⊢ φ → 2 ⁢ A ⁢ X − − B = − D ↔ - B + − D = 2 ⁢ A ⁢ X
106 88 6 subcld ⊢ φ → - B - D ∈ ℂ
107 106 10 5 96 divmuld ⊢ φ → - B - D 2 ⁢ A = X ↔ 2 ⁢ A ⁢ X = - B - D
108 eqcom ⊢ X = - B - D 2 ⁢ A ↔ - B - D 2 ⁢ A = X
109 eqcom ⊢ - B - D = 2 ⁢ A ⁢ X ↔ 2 ⁢ A ⁢ X = - B - D
110 107 108 109 3bitr4g ⊢ φ → X = - B - D 2 ⁢ A ↔ - B - D = 2 ⁢ A ⁢ X
111 103 105 110 3bitr4d ⊢ φ → 2 ⁢ A ⁢ X − − B = − D ↔ X = - B - D 2 ⁢ A
112 101 111 orbi12d ⊢ φ → 2 ⁢ A ⁢ X − − B = D ∨ 2 ⁢ A ⁢ X − − B = − D ↔ X = - B + D 2 ⁢ A ∨ X = - B - D 2 ⁢ A
113 87 91 112 3bitrd ⊢ φ → A ⁢ X 2 + B ⁢ X + C = 0 ↔ X = - B + D 2 ⁢ A ∨ X = - B - D 2 ⁢ A