Quadratic Solver Over Complexes

Math Algebra • Complex Numbers

Written by STEM Calculators Team Published December 20, 2025 Updated February 24, 2026

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3. Quadratic Solver Over Complexes

Solve \(ax^2+bx+c=0\) in \(\mathbb{C}\). Shows the discriminant and plots the roots on the complex plane.

Enable complex coefficients (a, b, c may have imaginary parts) Show step-by-step Compare real roots (when applicable)

Coefficient \(a\)

Re(a) Im(a)

Coefficient \(b\)

Re(b) Im(b)

Coefficient \(c\)

Re(c) Im(c)

Equal axis units (square grid) Label roots on plot Show conjugation hint (if real coefficients)

Ready

Drag to pan • wheel/pinch to zoom • Roots are plotted in the complex plane (Re horizontal, Im vertical).

Complex plane (Argand diagram)

Points correspond to \(x_1\) and \(x_2\).

Re: 0, Im: 0 sx: 70, sy: 70

Enter coefficients and click Solve.

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Theory

Quadratics over the complex numbers

A quadratic equation \[ ax^2 + bx + c = 0 \] can always be solved in the complex numbers \(\mathbb{C}\) as long as \(a\neq 0\). Even when there are no real solutions (for example when a real discriminant is negative), there are still complex solutions.

Discriminant

\[ \Delta = b^2 - 4ac. \]

Quadratic formula (valid in \(\mathbb{C}\))

\[ x = \frac{-b \pm \sqrt{\Delta}}{2a}. \]

If \(a,b,c\in\mathbb{R}\) and \(\Delta<0\), then the two roots form a conjugate pair: \[ x_2=\overline{x_1}. \] This conjugation symmetry is a property of real coefficients. With complex coefficients, roots do not need to be conjugates.

Complex square root (principal branch)

When \(\Delta\) is negative (or complex), \(\sqrt{\Delta}\) is taken as a complex square root. For a complex number \(z=u+iv\), one standard choice is the principal square root:

\[ \sqrt{z}=\alpha+i\beta,\qquad \alpha=\sqrt{\frac{|z|+u}{2}},\qquad \beta=\operatorname{sgn}(v)\sqrt{\frac{|z|-u}{2}}. \]

The calculator uses the principal square root so that the “\(+\)” and “\(-\)” in the quadratic formula produce the two roots consistently.

Vieta’s formulas (what the “Vieta check” means, even with imaginary parts)

Vieta’s formulas work over \(\mathbb{C}\) exactly as they do over \(\mathbb{R}\). If \(a\neq 0\) and the roots are \(x_1, x_2\in\mathbb{C}\), then:

\[ x_1+x_2=-\frac{b}{a},\qquad x_1x_2=\frac{c}{a}. \]

Why this is always true: the polynomial factors as

\[ ax^2+bx+c = a(x-x_1)(x-x_2) = a\Bigl(x^2-(x_1+x_2)x+x_1x_2\Bigr), \]

and matching coefficients gives the two identities above. No “real-only” assumption is used.

So what changes when imaginary parts are present? Not the correctness of Vieta — only what you should expect:

If \(a,b,c\in\mathbb{R}\), then \(-b/a\) and \(c/a\) are real, so \(x_1+x_2\) and \(x_1x_2\) are real. In the complex-root case, the roots are conjugates and appear symmetrically about the real axis.
If any coefficient is complex, then \(-b/a\) and \(c/a\) can be complex, and the roots are not required to be conjugates. The plot may have no symmetry about the real axis.

What the calculator’s Vieta check does: it compares the computed \(x_1+x_2\) with \(-b/a\), and \(x_1x_2\) with \(c/a\). Because calculations use floating-point arithmetic, the check is understood “up to a small numerical tolerance”.

Solved example

Solve \(x^2 + 2x + 5=0\).

\[ a=1,\quad b=2,\quad c=5. \]

\[ \Delta=b^2-4ac = 2^2 - 4\cdot 1\cdot 5 = 4 - 20 = -16. \]

\[ \sqrt{\Delta}=\sqrt{-16}=4i. \]

\[ x=\frac{-b\pm\sqrt{\Delta}}{2a} =\frac{-2\pm 4i}{2} = -1 \pm 2i. \]

Since \(a,b,c\) are real and \(\Delta<0\), the roots are complex conjugates (\(-1+2i\) and \(-1-2i\)) and appear symmetric about the real axis on the plot.

Graphical meaning

On the Argand diagram, the horizontal axis is \(\operatorname{Re}(x)\) and the vertical axis is \(\operatorname{Im}(x)\). Each root \(x\in\mathbb{C}\) is plotted as the point \((\operatorname{Re}(x),\operatorname{Im}(x))\).

Frequently Asked Questions

What does the discriminant mean for a quadratic solved over the complex numbers?

The discriminant Delta = b^2 - 4ac controls the square-root term in the quadratic formula. If coefficients are real and Delta is negative, the solutions are complex conjugates; with complex coefficients, Delta can be complex and the roots need not be conjugates.

How are complex roots computed from the quadratic formula?

The calculator evaluates x = (-b ± sqrt(Delta)) / (2a) using complex arithmetic. When Delta is complex, it computes a complex square root (typically the principal square root) so the ± choice consistently produces the two roots.

Why do complex roots come in conjugate pairs when a, b, and c are real?

Real coefficients imply the polynomial has conjugate symmetry: if x is a root, then its complex conjugate is also a root. This is why the two nonreal solutions appear symmetric about the real axis on the Argand diagram.

Can I solve a quadratic where a, b, or c has an imaginary part?

Yes. Enable complex coefficients and enter the real and imaginary parts for a, b, and c; the solver will compute roots in the complex plane even when the coefficients are not real.

How do I interpret the root plot on the complex plane?

Each root x = u + iv is plotted as the point (u, v), where u is the real part on the horizontal axis and v is the imaginary part on the vertical axis. If real coefficients produce conjugate roots, the points have equal real parts and opposite imaginary parts.

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Frequently Asked Questions

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