Raízes Equação Quadrática

Quadratic roots via discriminant

To solve a quadratic equation ax² + bx + c = 0 (assuming a ≠ 0), you use x = (-b ± √Δ)/(2a), with the discriminant given by Δ = b² - 4ac. Its sign tells you what kind of roots to expect. When Δ > 0 you get two distinct real roots; when Δ = 0 there's a single repeated real root; and when Δ < 0 the two roots are complex conjugates.

Example: take x² - 5x + 6 = 0. Here Δ = 25 - 24 = 1, so x = (5 ± 1)/2 and the roots are 3 and 2. You can double-check with Girard's relations: the sum S = -b/a = 5 and the product P = c/a = 6 both work out.

Applications

Quadratics describe projectile trajectories under constant gravity and turn up in geometric optics (parabolic mirrors and lenses), in engineering optimisation (areas, structural curves), and again and again in the ENEM and university entrance exams. The "Bhaskara formula" name is a Brazilian classroom tradition more than an accurate credit. Al-Khwarizmi (~820) was already solving quadratics geometrically, while Bhaskara II (12th c.) contributed mostly to related algebra and to negative roots.

FAQ

What if a = 0? Then it isn't a quadratic at all. The equation drops to linear form (bx + c = 0) and the discriminant has nothing to act on.

What does Δ mean geometrically? The parabola y = ax² + bx + c meets the x-axis exactly at the real roots. With Δ > 0 it crosses twice, with Δ = 0 it just touches, and with Δ < 0 it never reaches the axis.

What are complex conjugate roots? Whenever Δ < 0, the roots take the form x = -b/(2a) ± i·√(-Δ)/(2a). They arrive in pairs that share the same real part but carry opposite imaginary parts.

How do I factor using the roots? Call the roots r₁ and r₂. The polynomial then factors as ax² + bx + c = a(x - r₁)(x - r₂).