If α and β are solutions to the quadratic equation ax² + bx + c = 0, then we can substitute x = α and x = β into it

When x = α ,

aα² + bα + c = 0
aα² + bα = -c ①


When x = β ,

aβ² + bβ + c = 0
aβ² + bβ = -c ②

① = ②

aα² + bα = aβ² + bβ

aα² - aβ² = bβ - bα

a(α² - β²) = b(β - α)

a(α + β)(α - β) = -b(α - β)

a(α + β)(α - β) + b(α - β) = 0

(α - β)(a(α + β) + b) = 0

α - β = 0 or a(α + β) + b = 0

α = β or a(α + β) = -b

□□□□□ α + β = -b/a

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aα² + bα = -c
α(aα + b) = -c ③


aβ² + bβ = -c
β(aβ + b) = -c ④

③ x ④ :

α (aα + b) β (aβ + b) = c²

αβ (aα + b)(aβ + b) = c²

αβ (a²αβ + abα + abβ + b²) = c²

αβ (a²αβ + ab(α+β) + b²) = c²

αβ (a²αβ + ab(-b/a) + b²) = c²
(Based on earlier result)

αβ (a²αβ - b² + b²) = c²

αβ (a²αβ) = c²

(αβ)² a² = c²

(αβ)² = c² / a²

(αβ)² = (c/a)²

So αβ = c/a

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Alternatively,

① + ② :

aα² + bα + aβ² + bβ = -2c

a(α² + β²) + b(α + β) = -2c

a(α² + β²) + b(-b/a) = -2c

α² + β² - b²/a² = -2c / a

α² + β² - (-b/a)² = -2c / a

α² + β² - (α + β)² = -2c / a

α² + β² - (α² + 2αβ + β²) = -2c / a

-2αβ = -2c / a

αβ = c / a

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