Metamath Proof Explorer


Theorem 25or6to4

Description: Question 67 of 68 from a lecture Prof. Loof Lirpa held last Saturday in Lincoln Park. When asked why the smaller root wasn't reduced to 3/2, Lirpa responded "It really doesn't matter anyhow." (Contributed by Luke Murphy, 10-Jul-2026)

Ref Expression
Hypotheses 25or6to4.a φ A = 1
25or6to4.b φ B = 53 2
25or6to4.c φ C = 75 2
25or6to4.x φ X
Assertion 25or6to4 φ A X 2 + B X + C = 0 X = 25 X = 6 4

Proof

Step Hyp Ref Expression
1 25or6to4.a φ A = 1
2 25or6to4.b φ B = 53 2
3 25or6to4.c φ C = 75 2
4 25or6to4.x φ X
5 ax-1cn 1
6 1 5 eqeltrdi φ A
7 ax-1ne0 1 0
8 7 a1i φ 1 0
9 1 8 eqnetrd φ A 0
10 df-dec 53 = 9 + 1 5 + 3
11 9cn 9
12 11 5 addcli 9 + 1
13 5cn 5
14 12 13 mulcli 9 + 1 5
15 3cn 3
16 14 15 addcli 9 + 1 5 + 3
17 10 16 eqeltri 53
18 2cn 2
19 2ne0 2 0
20 17 18 19 divcli 53 2
21 20 negcli 53 2
22 2 21 eqeltrdi φ B
23 df-dec 75 = 9 + 1 7 + 5
24 7cn 7
25 12 24 mulcli 9 + 1 7
26 25 13 addcli 9 + 1 7 + 5
27 23 26 eqeltri 75
28 27 18 19 divcli 75 2
29 3 28 eqeltrdi φ C
30 df-dec 25 = 9 + 1 2 + 5
31 12 18 mulcli 9 + 1 2
32 31 13 addcli 9 + 1 2 + 5
33 30 32 eqeltri 25
34 33 a1i φ 25
35 6cn 6
36 4cn 4
37 4ne0 4 0
38 35 36 37 divcli 6 4
39 38 a1i φ 6 4
40 33 18 19 divcan4i 25 2 2 = 25
41 2nn0 2 0
42 5nn0 5 0
43 eqid 25 = 25
44 0nn0 0 0
45 1nn0 1 0
46 2t2e4 2 2 = 4
47 46 oveq1i 2 2 + 1 = 4 + 1
48 4p1e5 4 + 1 = 5
49 47 48 eqtri 2 2 + 1 = 5
50 5t2e10 5 2 = 10
51 41 41 42 43 44 45 49 50 decmul1c 25 2 = 50
52 51 oveq1i 25 2 2 = 50 2
53 40 52 eqtr3i 25 = 50 2
54 6t2e12 6 2 = 12
55 4t3e12 4 3 = 12
56 36 15 55 mulcomli 3 4 = 12
57 54 56 eqtr4i 6 2 = 3 4
58 35 15 pm3.2i 6 3
59 36 37 pm3.2i 4 4 0
60 2cnne0 2 2 0
61 59 60 pm3.2i 4 4 0 2 2 0
62 58 61 pm3.2i 6 3 4 4 0 2 2 0
63 divmuleq 6 3 4 4 0 2 2 0 6 4 = 3 2 6 2 = 3 4
64 62 63 ax-mp 6 4 = 3 2 6 2 = 3 4
65 57 64 mpbir 6 4 = 3 2
66 53 65 oveq12i 25 + 6 4 = 50 2 + 3 2
67 dfdec10 53 = 10 5 + 3
68 42 dec0u 10 5 = 50
69 68 oveq1i 10 5 + 3 = 50 + 3
70 67 69 eqtri 53 = 50 + 3
71 70 oveq1i 53 2 = 50 + 3 2
72 df-dec 50 = 9 + 1 5 + 0
73 0cn 0
74 14 73 addcli 9 + 1 5 + 0
75 72 74 eqeltri 50
76 75 15 18 19 divdiri 50 + 3 2 = 50 2 + 3 2
77 71 76 eqtri 53 2 = 50 2 + 3 2
78 66 77 eqtr4i 25 + 6 4 = 53 2
79 78 20 eqeltri 25 + 6 4
80 79 negnegi 25 + 6 4 = 25 + 6 4
81 78 negeqi 25 + 6 4 = 53 2
82 21 div1i 53 2 1 = 53 2
83 81 82 eqtr4i 25 + 6 4 = 53 2 1
84 83 negeqi 25 + 6 4 = 53 2 1
85 80 84 eqtr3i 25 + 6 4 = 53 2 1
86 2 1 oveq12d φ B A = 53 2 1
87 86 negeqd φ B A = 53 2 1
88 85 87 eqtr4id φ 25 + 6 4 = B A
89 28 div1i 75 2 1 = 75 2
90 33 15 18 19 divassi 25 3 2 = 25 3 2
91 3nn0 3 0
92 2t3e6 2 3 = 6
93 92 oveq1i 2 3 + 1 = 6 + 1
94 6p1e7 6 + 1 = 7
95 93 94 eqtri 2 3 + 1 = 7
96 5t3e15 5 3 = 15
97 91 41 42 43 42 45 95 96 decmul1c 25 3 = 75
98 97 oveq1i 25 3 2 = 75 2
99 90 98 eqtr3i 25 3 2 = 75 2
100 65 oveq2i 25 6 4 = 25 3 2
101 100 eqcomi 25 3 2 = 25 6 4
102 89 99 101 3eqtr2ri 25 6 4 = 75 2 1
103 3 1 oveq12d φ C A = 75 2 1
104 102 103 eqtr4id φ 25 6 4 = C A
105 6 9 22 29 4 34 39 88 104 quadfac φ A X 2 + B X + C = 0 X = 25 X = 6 4