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Solve using difference of squares

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a+bz+cz^2=0

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a + bz + cz² = 0 cz² + bz + a = 0 z² + (b/c).z + (a/c) = 0 z² + (b/c).z + (b/2c)² - (b/2c)² + (a/c) = 0 [z² + (b/c).z + (b/2c)²] - [(b/2c)² - (a/c)] = 0 [z + (b/2c)]² - [(b/2c)² - (a/c)] = 0 [z + (b/2c)]² - [(b²/4c²) - (4ac/4c²)] = 0 [z + (b/2c)]² - [(b² - 4ac)/4c²] = 0 → let: Δ = b² - 4ac [z + (b/2c)]² - [Δ/4c²] = 0 [z + (b/2c)]² - [(√Δ)/2c]² = 0 → you recognize the difference between 2 squares: a² - b² = (a + b).(a - b) [z + (b/2c) + (√Δ)/2c].[z + (b/2c) - (√Δ)/2c] = 0 [z + (b + √Δ)/2c].[z + (b - √Δ)/2c] = 0 [z - (- b - √Δ)/2c].[z - (- b + √Δ)/2c] = 0 → let: z₁ = (- b - √Δ)/2c (z - z₁).[z - (- b + √Δ)/2c] = 0 → let: z₂ = (- b + √Δ)/2c (z - z₁).(z - z₂) = 0 → you can see that the roots are: z₁ = (- b - √Δ)/2c z₂ = (- b + √Δ)/2c

            
        

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