x² + 1 = 0
- Input
- a=1, b=0, c=1
- Expected output
- x₁ = 0 + 1i, x₂ = 0 − 1i
Δ = 0² − 4×1×1 = −4; real part 0, imaginary part √4/2 = 1.
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complex roots quadratic equation
When the discriminant Δ = b² − 4ac is negative, ax² + bx + c = 0 has no real solution: the parabola never crosses the x-axis. Both roots exist only among complex numbers, in the form x = (−b ± i√|Δ|) / (2a), where i is the imaginary unit (i² = −1).
Δ = 0² − 4×1×1 = −4; real part 0, imaginary part √4/2 = 1.
Δ = 2² − 4×1×5 = −16; the roots sum to −2 = −b/a, confirming Vieta's formula.
Δ = 2² − 4×2×5 = −36; each root's modulus is √(0.5² + 1.5²) ≈ 1.58.
The discriminant Δ = b²−4ac determines the nature of roots: Δ>0 → two distinct real roots; Δ=0 → double root; Δ<0 → no real roots.
Yes. The step-by-step first computes Δ = b² − 4ac; whenever Δ < 0, the tool automatically switches from real x₁/x₂ to the a ± bi format, with no extra setting required.
Add the two roots together: the result should equal exactly −b/a. Multiply them together: the result should equal exactly c/a. Both checks (Vieta's formulas) work even when the roots are not real numbers.
No. The formula x = (−b ± i√|Δ|)/(2a) is the same for any real a, b, c with a≠0; only the numeric result changes, as in 2x²+2x+5=0, whose roots are −0.5 ± 1.5i.
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