Quadratic Equation Solver guide
How to Solve a Quadratic with Complex Roots
When the discriminant is negative, take the square root of its absolute value and attach i, the imaginary unit. For x² + 4x + 13 = 0 the discriminant is −36, and √−36 = 6i, giving roots of x = −2 + 3i and x = −2 − 3i.
A negative discriminant is not an error
The first time the square root comes out negative, it feels like something has gone wrong. It usually has not.
It means the parabola never crosses the x-axis, so there are no real solutions. But there are still two solutions — they just live in the complex numbers.
If your course has not covered complex numbers yet, the correct answer at this point is "no real solutions", and that is a complete answer. If it has, read on.
What i means
i is defined as the square root of −1. That is the whole definition.
It exists because no real number squares to give a negative — any real number times itself is positive or zero. So mathematicians defined a new one that does.
From that single definition everything else follows. √−36 = √36 × √−1 = 6i. √−9 = 3i. √−5 = i√5.
A worked example
Take x² + 4x + 13 = 0, so a = 1, b = 4, c = 13.
Discriminant: 4² − 4(1)(13) = 16 − 52 = −36. Negative, so the roots are complex.
√−36 = 6i.
Substituting: x = (−4 ± 6i) ÷ 2.
Divide both parts by 2: x = −2 ± 3i.
So the roots are x = −2 + 3i and x = −2 − 3i.
Dividing both parts
That last division is where most errors creep in. It is easy to divide the real part and forget the imaginary one, giving −2 ± 6i.
Both terms are on the top of the fraction, so both get divided. (−4 + 6i) ÷ 2 is −2 + 3i, exactly as (4 + 6) ÷ 2 would be 2 + 3.
Writing the fraction out with both terms visible before dividing makes this much harder to get wrong.
Complex roots always come in pairs
Notice that the two roots are identical apart from the sign in the middle: −2 + 3i and −2 − 3i. That pairing is called a complex conjugate pair.
It is not a coincidence of this example. For any quadratic with real coefficients, complex roots always arrive in conjugate pairs — the ± in the formula guarantees it.
So if you find one complex root, you already know the other. And if a question gives you one and asks for the second, flipping the sign of the imaginary part is the entire answer.
Checking a complex root
The check works the same as with real roots, remembering that i² = −1.
Substituting x = −2 + 3i into x² + 4x + 13:
(−2 + 3i)² = 4 − 12i + 9i² = 4 − 12i − 9 = −5 − 12i.
4(−2 + 3i) = −8 + 12i.
Adding it all: (−5 − 12i) + (−8 + 12i) + 13 = 0. The imaginary parts cancel, and the real parts come to zero.
That cancellation is a good sign the working is right — it should always happen.
