Metamath Proof Explorer


Theorem quad1

Description: A condition for a quadratic equation with complex coefficients to have (exactly) one complex solution. (Contributed by AV, 23-Jan-2023)

Ref Expression
Hypotheses quad1.a φ A
quad1.z φ A 0
quad1.b φ B
quad1.c φ C
quad1.d φ D = B 2 4 A C
Assertion quad1 φ ∃! x A x 2 + B x + C = 0 D = 0

Proof

Step Hyp Ref Expression
1 quad1.a φ A
2 quad1.z φ A 0
3 quad1.b φ B
4 quad1.c φ C
5 quad1.d φ D = B 2 4 A C
6 1 adantr φ x A
7 2 adantr φ x A 0
8 3 adantr φ x B
9 4 adantr φ x C
10 simpr φ x x
11 5 adantr φ x D = B 2 4 A C
12 6 7 8 9 10 11 quad φ x A x 2 + B x + C = 0 x = - B + D 2 A x = - B - D 2 A
13 12 reubidva φ ∃! x A x 2 + B x + C = 0 ∃! x x = - B + D 2 A x = - B - D 2 A
14 3 negcld φ B
15 3 sqcld φ B 2
16 4cn 4
17 16 a1i φ 4
18 1 4 mulcld φ A C
19 17 18 mulcld φ 4 A C
20 15 19 subcld φ B 2 4 A C
21 5 20 eqeltrd φ D
22 21 sqrtcld φ D
23 14 22 addcld φ - B + D
24 2cnd φ 2
25 24 1 mulcld φ 2 A
26 2ne0 2 0
27 26 a1i φ 2 0
28 24 1 27 2 mulne0d φ 2 A 0
29 23 25 28 divcld φ - B + D 2 A
30 14 22 subcld φ - B - D
31 30 25 28 divcld φ - B - D 2 A
32 euoreqb - B + D 2 A - B - D 2 A ∃! x x = - B + D 2 A x = - B - D 2 A - B + D 2 A = - B - D 2 A
33 29 31 32 syl2anc φ ∃! x x = - B + D 2 A x = - B - D 2 A - B + D 2 A = - B - D 2 A
34 14 22 25 28 divdird φ - B + D 2 A = B 2 A + D 2 A
35 14 22 25 28 divsubdird φ - B - D 2 A = B 2 A D 2 A
36 14 25 28 divcld φ B 2 A
37 22 25 28 divcld φ D 2 A
38 36 37 negsubd φ B 2 A + D 2 A = B 2 A D 2 A
39 22 25 28 divnegd φ D 2 A = D 2 A
40 39 oveq2d φ B 2 A + D 2 A = B 2 A + D 2 A
41 35 38 40 3eqtr2d φ - B - D 2 A = B 2 A + D 2 A
42 34 41 eqeq12d φ - B + D 2 A = - B - D 2 A B 2 A + D 2 A = B 2 A + D 2 A
43 22 negcld φ D
44 43 25 28 divcld φ D 2 A
45 36 37 44 addcand φ B 2 A + D 2 A = B 2 A + D 2 A D 2 A = D 2 A
46 div11 D D 2 A 2 A 0 D 2 A = D 2 A D = D
47 22 43 25 28 46 syl112anc φ D 2 A = D 2 A D = D
48 22 eqnegd φ D = D D = 0
49 cnsqrt00 D D = 0 D = 0
50 21 49 syl φ D = 0 D = 0
51 48 50 bitrd φ D = D D = 0
52 45 47 51 3bitrd φ B 2 A + D 2 A = B 2 A + D 2 A D = 0
53 42 52 bitrd φ - B + D 2 A = - B - D 2 A D = 0
54 13 33 53 3bitrd φ ∃! x A x 2 + B x + C = 0 D = 0