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)u − v means u + (−v), and b/a means b·a⁻¹.convention, A12
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Lemma L1
For all p, α, β, p(x − α)(x − β) = px² − p(α + β)x + pαβ
Proof.
(1)(x − α)(x − β) = x(x − β) + (−α)(x − β)A8, A10
(2)x(x − β) = x² + (−(βx))A8, A18, A10
(3)(−α)(x − β) = (−(αx)) + αβA8, A18, A21
(4)(x − α)(x − β) = x² + (−(βx)) + (−(αx)) + αβ(1), (2), (3), A13, A14
(5)(−(βx)) + (−(αx)) = −(βx + αx)A19
(6)βx + αx = (β + α)x = (α + β)xA8, A10, A14
(7)(x − α)(x − β) = x² − (α + β)x + αβ(4), (5), (6)
(8)Multiplying both sides by p gives p(x−α)(x−β) = px² + p(−((α+β)x)) + p(αβ)(7), E, A8, A13, A9
(9)p(−((α+β)x)) = −(p(α+β)x)A18, A10, A9
(10)p(x−α)(x−β) = px² − p(α+β)x + pαβ(8), (9), A9
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Lemma L2
If distinct α, β are roots of ax² + bx + c = 0, then there is a p with ax² + bx + c = p(x − α)(x − β).
Proof.
(1)Set f(x) = ax² + bx + c.convention
(2)Since f(α) = 0, there are p, d with f(x) = (x − α)(px + d).assumption, D1, A1
(3)Substituting x = β gives 0 = (β − α)(pβ + d)(2), assumption, D1, A3, A4, A5, A6
(4)Since β ≠ α, we have β − α ≠ 0.assumption, E, A13, A14, A15, A16
(5)The product is 0, so pβ + d = 0.(3), (4), A7
(6)Adding −(pβ) to both sides gives d = −(pβ)(5), E, A13, A14, A15, A16
(7)px + d = px + (−(pβ)) = p(x − β)(6), A8, A18, A10
(8)f(x) = (x − α)·p(x − β) = p(x − α)(x − β)(2), (7), A9, A10
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Theorem T
If α, β are two distinct roots of ax² + bx + c = 0, then α + β = −b/a, αβ = c/a
Proof.
(1)There is a p with f(x) = p(x − α)(x − β).L2
(2)p(x−α)(x−β) = px² − p(α+β)x + pαβL1
(3)For every x, ax² + bx + c = px² − p(α+β)x + pαβ.(1), (2)
(4)Comparing coefficients gives a = p, b = −(a(α+β)), c = aαβ.(3), A2
(5)b + a(α + β) = 0(4), E, A14, A16
(6)Multiplying both sides by a⁻¹ gives a⁻¹b + (α + β) = 0(5), E, A8, A9, A12, A10, A11, A17
(7)Since a⁻¹b + (−(a⁻¹b)) = 0, we get α + β = −(a⁻¹b) = −b/a(6), A16, A20, A10, D1
(8)Multiplying both sides of c = aαβ by a⁻¹ gives a⁻¹c = (a⁻¹a)αβ = αβ(4), E, A9, A12, A10, A11
(9)αβ = ca⁻¹ = c/a(8), A10, D1
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Corollary C
For any α, β, the only roots of x² − (α + β)x + αβ = 0 are α and β.
Proof.
(1)Setting p = 1 in L1 gives (x − α)(x − β) = x² − (α+β)x + αβL1, A11, A10
(2)Substituting x = α gives (α − α)(α − β) = 0 × (α − β) = 0(1), A3, A4, A5, A6, A16, A17, A10
(3)Substituting x = β gives (β − α)(β − β) = (β − α) × 0 = 0(1), A3, A4, A5, A6, A16, A17
(4)Hence α and β are roots.(2), (3), D1
(5)If γ is a root, then (γ − α)(γ − β) = 0.(1), D1, A3, A4, A5, A6
(6)If γ ≠ α and γ ≠ β, then γ − α ≠ 0, γ − β ≠ 0, so the product is not 0.E, A13, A14, A15, A16, A7
(7)Hence the only roots are α and β.(5), (6)
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