Definition D1
A root of ax² + bx + c = 0 (a ≠ 0) is a number x for which the equation holds upon substitution.
Proof.
(1)a ≠ 0 is the condition for this expression to be quadratic, and is assumed in every block below.convention
(2)The number of roots means the number of all such x.convention
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Definition D2
D denotes b² − 4ac.
Proof.
This value is called the discriminant.convention
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Lemma L1
When a ≠ 0, for each number u there is exactly one x with 2ax + b = u.
Proof.
(1)Since 2 ≠ 0 and a ≠ 0, we have 2a ≠ 0.A17, A16
(2)Set x₀ = (u + (−b))(2a)⁻¹.(1), A14
(3)2ax₀ = (2a(2a)⁻¹)(u + (−b)) = u + (−b)(2), A11, A12, A14, A13
(4)2ax₀ + b = (u + (−b)) + b = u(3), A7, A8, A10, A9
(5)Suppose 2ax + b = u and 2ay + b = u.assumption
(6)Adding −b to both sides gives 2ax = 2ay.(5), E, A7, A10, A9
(7)Multiplying both sides by (2a)⁻¹ gives x = y.(6), (1), E, A11, A12, A14, A13
(8)Such an x exists and is unique.(4), (7)
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Theorem T1
If D > 0, there are exactly two roots.
Proof.
(1)Suppose D > 0.assumption
(2)There exists a unique s > 0 with s² = D.(1), A2
(3)ax² + bx + c = 0 ⟺ (2ax + b)² = D = s²A1, D2, (2)
(4)(2ax + b)² = s² ⟺ 2ax + b = s or 2ax + b = −sA3
(5)Since s > 0, we have s ≠ 0.(2), A5
(6)Since 2 ≠ 0, we have 2s ≠ 0.(5), A17, A16
(7)If s = −s, then s + s = s + (−s) = 0.A10
(8)s + s = s(1 + 1) = 2sA15, A13, A12
(9)Hence s = −s contradicts (6), and s ≠ −s.(6), (7), (8)
(10)There is exactly one x with 2ax + b = s and exactly one x with 2ax + b = −s.L1
(11)If the two coincided, then s = −s, contradicting (9).(9), (10)
(12)There are exactly two roots.(3), (4), (10), (11), D1
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Theorem T2
If D = 0, there is exactly one root.
Proof.
(1)Suppose D = 0.assumption
(2)ax² + bx + c = 0 ⟺ (2ax + b)² = 0A1, D2, (1)
(3)If u ≠ 0, then u² ≠ 0.A16
(4)Hence if (2ax + b)² = 0, then 2ax + b = 0.(3)
(5)Conversely, if 2ax + b = 0, then (2ax + b)² = 0 × 0 = 0.A18
(6)There is exactly one x with 2ax + b = 0.L1
(7)There is exactly one root.(2), (4), (5), (6), D1
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Theorem T3
If D < 0, no real root exists.
Proof.
(1)Suppose D < 0.assumption
(2)If a real root x existed, then (2ax + b)² = D.A1, D1, D2
(3)0 ≤ (2ax + b)²A4
(4)Hence 0 ≤ D.(2), (3), E
(5)Trichotomy does not allow D < 0 and 0 ≤ D together.(1), (4), A5
(6)No such x exists.(5)
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Corollary C
Even when D < 0, on the stage extended by i there are exactly two roots.
Proof.
(1)Suppose D < 0.assumption
(2)A1, A3, L1 use only the field axioms, so they hold over the complex numbers as well.A1, A3, L1
(3)Adding −D to both sides gives 0 < −D.(1), A6, A10, A8, A9
(4)There exists s > 0 with s² = −D.(3), A2
(5)There is i ≠ 0 with i² = −1.A19
(6)(is)² = i²s² = (−1)s²(5), A11, A12, E
(7)(−1)s² = (−1)(−D) = 1 × D = D(4), A20, A12, A13, E
(8)(2ax + b)² = D = (is)², so 2ax + b = ±is.(2), (6), (7), A3
(9)Since s ≠ 0, i ≠ 0, 2 ≠ 0, we have 2is ≠ 0.(4), (5), A5, A17, A16
(10)If is = −is, then is + is = 2(is) = 0, contradicting (9).(9), A10, A15, A13, A12
(11)There is exactly one x with 2ax + b = is and exactly one x with 2ax + b = −is, and they differ.(2), (10), L1
(12)There are exactly two roots.(8), (11), D1
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