Complex Roots of Quadratics

Intuition

The Quadratic Formula lesson ended with an honest refusal. When the discriminant $\Delta = b^2 - 4ac$ came out negative, the parabola floated clear of the $x$ -axis and the formula asked for the square root of a negative number. In that case the only correct answer available was that no real solution exists. That refusal was never a fact about the equation. It was a fact about the numbers we were allowed to use. This chapter built the missing numbers. You now own a number $i$ with $i^2 = -1$ , arithmetic for numbers of the form $a + bi$ , and a conjugate operation that flips the sign of the imaginary part. This closing lesson brings those tools back to the question that motivated the whole chapter. Run the quadratic formula when $\Delta < 0$ and do not stop. Every quadratic turns out to have exactly two roots after all. The pair the discriminant said were missing were never missing; they sit just off the real line, mirror images of each other across it.

Learning goals

Convert to $i$ -form before any radical arithmetic
Run the quadratic formula through a negative discriminant
Prove the conjugate root theorem for real coefficients
Read the shared real part as the axis of symmetry
Rebuild the monic quadratic from one non-real root
Factor every quadratic over the complex numbers

Square roots of negative numbers, pinned down

Everything hinges on one question the real numbers could not answer: what number squares to $-16$ ? Inside the reals, nothing does, because a real square is never negative. But with $i$ in hand the question has an answer, and you can check it by squaring. For any positive real $k$ ,

\left(i\sqrt{k}\right)^2 = i^2\left(\sqrt{k}\right)^2 = (-1)\,k = -k, \qquad \left(-i\sqrt{k}\right)^2 = (-i)^2 k = i^2 k = -k.

So a negative number $-k$ has exactly two square roots, $i\sqrt{k}$ and $-i\sqrt{k}$ , both pure imaginary. Just as $\sqrt{9}$ names the non-negative choice $3$ rather than $-3$ , we adopt the convention that the symbol $\sqrt{-k}$ names the choice $i\sqrt{k}$ :

\sqrt{-k} = i\sqrt{k} \qquad \text{for } k > 0, \qquad \text{for example } \sqrt{-16} = 4i.

The other square root, $-4i$ , is not lost; the $\pm$ in the quadratic formula will supply it, exactly as it supplied $-3$ alongside $3$ in the real case.

One habit must be installed before anything else, because it prevents the single most common error in this topic. The familiar radical rule $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ was proved for non-negative $a$ and $b$ , and it fails when both are negative. Convert to $i$ -form first, then multiply:

\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6, \qquad \text{not} \qquad \sqrt{(-4)(-9)} = \sqrt{36} = 6.

The safe order of operations is always the same: rewrite every $\sqrt{-k}$ as $i\sqrt{k}$ at the moment it appears, and only then do arithmetic with the results.

Check your understanding

Rewrite $\sqrt{-45}$ in terms of $i$ , in simplest form.

3i\sqrt{5}

9i\sqrt{5}

-3\sqrt{5}

5i\sqrt{3}

The formula runs to completion

Return to the derivation of the quadratic formula. Completing the square on $ax^2 + bx + c = 0$ reached the line

\left(x + \frac{b}{2a}\right)^2 = \frac{\Delta}{4a^2}, \qquad \Delta = b^2 - 4ac,

and in Chapter 4 the case $\Delta < 0$ stopped there: no real square could equal a negative number. Now nothing stops.

A negative discriminant produces the pair $p \pm qi$ #

Suppose $a$ , $b$ , $c$ are real, $a \ne 0$ , and $\Delta = b^2 - 4ac < 0$ . Write $|\Delta| = 4ac - b^2$ , which is a positive real number, so the right side above is the negative real $-\dfrac{|\Delta|}{4a^2}$ .

By the opening section, a negative real has exactly two square roots, and here they are $\pm\, i\,\dfrac{\sqrt{|\Delta|}}{2|a|}$ , since squaring either one gives $-\dfrac{|\Delta|}{4a^2}$ . The $\pm$ already supplies both signs, so replacing $2|a|$ by $2a$ merely trades which sign is which, exactly as in the real derivation. Therefore

x + \frac{b}{2a} = \pm\, i\,\frac{\sqrt{4ac - b^2}}{2a},

and subtracting $\tfrac{b}{2a}$ gives the two solutions

x = -\frac{b}{2a} \;\pm\; \frac{\sqrt{4ac - b^2}}{2a}\, i.

Both fractions are real numbers: the radicand $4ac - b^2$ is positive, so its real square root exists. Name them

p = -\frac{b}{2a}, \qquad q = \frac{\sqrt{4ac - b^2}}{2a},

and the two roots are exactly $p + qi$ and $p - qi$ , with $q \ne 0$ because the radicand is strictly positive. In practice you do not re-derive this: you run the ordinary quadratic formula and read $\sqrt{\Delta}$ as $i\sqrt{|\Delta|}$ the moment $\Delta$ is negative. The formula never needed changing; it needed numbers that could finish it.

Two features of the answer deserve a hard look before any examples. First, the real part $p = -\tfrac{b}{2a}$ is the axis of symmetry, the same axis the real roots straddled in Chapter 4. The roots still straddle it; the offset $\pm q$ has simply rotated off the real line into the imaginary direction. Second, the two roots have the same real part and opposite imaginary parts, so each is the conjugate of the other. That is not a small remark, and the next section proves it is no accident of the formula.

Worked example 1 The discriminant is negative; solve anyway

Solve $x^2 - 6x + 13 = 0$ .

Here $a = 1$ , $b = -6$ , $c = 13$ , so the discriminant is

\Delta = (-6)^2 - 4(1)(13) = 36 - 52 = -16.

In Chapter 4 this was the end: no real solutions. Now read $\sqrt{-16} = 4i$ and keep going:

x = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i.

The two roots are $3 + 2i$ and $3 - 2i$ , a conjugate pair with real part on the axis $x = 3$ . A root this surprising deserves a direct check. Substitute $x = 3 + 2i$ , using $(3 + 2i)^2 = 9 + 12i + 4i^2 = 5 + 12i$ :

(5 + 12i) - 6(3 + 2i) + 13 = 5 + 12i - 18 - 12i + 13 = 0.

It works, with the imaginary parts cancelling on their own. The conjugate $3 - 2i$ checks the same way, with every $i$ replaced by $-i$ throughout.

Worked example 2 A leading coefficient and fraction roots

Solve $4x^2 - 8x + 13 = 0$ .

With $a = 4$ , $b = -8$ , $c = 13$ ,

\Delta = (-8)^2 - 4(4)(13) = 64 - 208 = -144, \qquad \sqrt{-144} = 12i.

Now apply the formula, and divide both terms of the numerator by the full denominator $2a = 8$ :

x = \frac{8 \pm 12i}{8} = 1 \pm \frac{3}{2}\,i.

Here $p = 1$ and $q = \tfrac{3}{2}$ . Sanity checks are cheap and worth the seconds: the real part $1$ should be the axis $-\tfrac{b}{2a} = \tfrac{8}{8} = 1$ , and it is. Leaving the answer as $\tfrac{8 \pm 12i}{8}$ unreduced, or dividing only the $8$ and not the $12i$ , are the two standard ways this problem goes wrong.

Check your understanding

Solve $x^2 + 4x + 5 = 0$ .

x = -2 \pm i

x = 2 \pm i

x = -2 \pm 2i

x = -4 \pm 2i

Worked example 3 An irrational imaginary part

Solve $x^2 + 3x + 3 = 0$ .

The discriminant is $\Delta = 3^2 - 4(1)(3) = 9 - 12 = -3$ , so $\sqrt{-3} = i\sqrt{3}$ and

x = \frac{-3 \pm i\sqrt{3}}{2} = -\frac{3}{2} \pm \frac{\sqrt{3}}{2}\,i.

Both forms name the same pair; the second displays the real and imaginary parts $p = -\tfrac{3}{2}$ and $q = \tfrac{\sqrt{3}}{2}$ explicitly. Nothing requires $q$ to be a whole number or even rational: the imaginary part inherits whatever radical $\sqrt{|\Delta|}$ leaves behind, exactly as irrational real roots did in Chapter 4.

Non-real roots always arrive in conjugate pairs

The formula handed us conjugate pairs, but that could in principle be a habit of the formula rather than a law of the equations. It is a law. Real coefficients alone force it, by an argument that never mentions the quadratic formula and will survive, word for word, for polynomials of every degree.

The conjugate root theorem for quadratics#

Let $a$ , $b$ , $c$ be real with $a \ne 0$ , and suppose the complex number $z$ satisfies $az^2 + bz + c = 0$ . We claim $\overline{z}$ satisfies it too.

The proof rests on how conjugation interacts with arithmetic, so verify that first. Write $w = p + qi$ and $v = r + si$ with $p, q, r, s$ real. For addition, $w + v = (p + r) + (q + s)i$ , so

\overline{w + v} = (p + r) - (q + s)i = (p - qi) + (r - si) = \overline{w} + \overline{v}.

For multiplication, expanding $wv$ gives $wv = (pr - qs) + (ps + qr)i$ , so $\overline{wv} = (pr - qs) - (ps + qr)i$ ; and multiplying the conjugates directly,

\overline{w}\,\overline{v} = (p - qi)(r - si) = pr - psi - qri + qs\,i^2 = (pr - qs) - (ps + qr)i,

the same number. So conjugation passes through sums and through products, and in particular $\overline{z^2} = \overline{z}^{\,2}$ . Finally, a real number $r = r + 0i$ is its own conjugate.

Now conjugate the whole equation $az^2 + bz + c = 0$ . The right side is real, so it conjugates to itself, $\overline{0} = 0$ . The left side is a sum of products, so the conjugate passes through every join, and the real coefficients $a$ , $b$ , $c$ come through unchanged:

0 = \overline{az^2 + bz + c} = \overline{a}\,\overline{z}^{\,2} + \overline{b}\,\overline{z} + \overline{c} = a\overline{z}^{\,2} + b\overline{z} + c.

That last expression is the original quadratic evaluated at $\overline{z}$ , and it equals $0$ , so $\overline{z}$ is a root.

Read the theorem’s fine print, because both halves of it matter. When $z$ is real, $\overline{z} = z$ and the statement says nothing new; its full force is for non-real roots, which it forbids from ever appearing alone. A quadratic with real coefficients therefore has exactly three possible root inventories: two distinct real roots, one repeated real root, or a conjugate pair off the real line. “One real root and one non-real root” is not on the list, for any real coefficients whatsoever. And the hypothesis is not decoration: the quadratic $x^2 - 2ix - 1 = (x - i)^2$ has the double root $i$ , and its conjugate $-i$ is not a root at all. No contradiction, because the coefficient $-2i$ is not real. State the theorem with its hypothesis attached, every time.

The same equation seen twice. On the real line the parabola with negative discriminant never meets the x-axis, so no real root exists. In the complex plane the two roots are there all along, sitting at p + qi and p - qi, mirror images of each other across the real axis.

Sum and product still tell the truth

Chapter 4 closed with the relations $r_1 + r_2 = -\tfrac{b}{a}$ and $r_1 r_2 = \tfrac{c}{a}$ , but the proof there leaned on a factorization over the reals, which assumed the roots were real. Do the relations survive the move off the real line? They do, and the cleanest way to see it is to compute both quantities straight from the pair $p \pm qi$ . The sum is immediate, because the imaginary parts cancel:

(p + qi) + (p - qi) = 2p = 2\left(-\frac{b}{2a}\right) = -\frac{b}{a}.

The product is a conjugate product, the difference of squares in disguise:

(p + qi)(p - qi) = p^2 - (qi)^2 = p^2 - q^2 i^2 = p^2 + q^2,

and substituting the formulas for $p$ and $q$ collapses it to a single fraction:

p^2 + q^2 = \frac{b^2}{4a^2} + \frac{4ac - b^2}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}.

Both relations hold verbatim, so everything you learned to do with sums and products of roots still works when the roots are complex. Two extra facts fall out of the product for free. First, $p^2 + q^2$ is a sum of real squares with $q \ne 0$ , so it is strictly positive. Therefore, whenever a real quadratic has non-real roots, $\tfrac{c}{a} > 0$ , meaning $a$ and $c$ must share a sign. That agrees with the discriminant, since $b^2 < 4ac$ forces $ac > \tfrac{b^2}{4} \ge 0$ . Second, the previous lesson measured a complex number’s distance from the origin by its modulus, with $|z|^2 = p^2 + q^2$ . So the product of a conjugate pair is exactly the squared modulus of either root:

r_1 r_2 = |z|^2 = \frac{c}{a}.

For a monic quadratic $x^2 + bx + c$ with non-real roots, each root sits at distance $\sqrt{c}$ from the origin of the complex plane. The coefficients were carrying geometry the whole time.

The practical payoff is reconstruction. One non-real root of a real quadratic determines everything, because the conjugate theorem hands you the other root, and then the sum and product hand you the coefficients. A monic quadratic with roots $p \pm qi$ must be

x^2 - (\text{sum})\,x + (\text{product}) = x^2 - 2p\,x + \left(p^2 + q^2\right).

Worked example 4 Rebuild the quadratic from one root

A monic quadratic with real coefficients has $2 + i\sqrt{5}$ as one of its roots. Find the quadratic.

Real coefficients force the other root to be the conjugate, $2 - i\sqrt{5}$ . Now compute the sum and the product. The sum is $2p = 4$ . The product is the squared modulus:

\left(2 + i\sqrt{5}\right)\left(2 - i\sqrt{5}\right) = 2^2 + \left(\sqrt{5}\right)^2 = 4 + 5 = 9.

The monic quadratic with that sum and product is

x^2 - 4x + 9 = 0.

Verify by solving it: $\Delta = 16 - 36 = -20$ , so $x = \tfrac{4 \pm 2i\sqrt{5}}{2} = 2 \pm i\sqrt{5}$ , recovering the given root and its conjugate. One complex root pinned down the entire equation, up to an overall scalar.

Check your understanding

A monic quadratic with real coefficients has $6 + i$ as one root. Which equation is it?

x^2 - 12x + 37 = 0

x^2 + 12x + 37 = 0

x^2 - 12x + 35 = 0

x^2 - 6x + 37 = 0

Every quadratic factors completely now

The factoring lesson in Chapter 4 insisted that “factors” is meaningless until you name the number system. Over the reals a quadratic with $\Delta < 0$ is genuinely irreducible: no real roots, so no real linear factors. Over the complex numbers the last holdouts fall. If $z$ and $\overline{z}$ are the roots, then

\begin{aligned} a(x - z)(x - \overline{z}) &= a\left(x^2 - \left(z + \overline{z}\right)x + z\overline{z}\right) \\ &= a\left(x^2 - \left(-\frac{b}{a}\right)x + \frac{c}{a}\right) = ax^2 + bx + c, \end{aligned}

using the sum and product just proved. So every quadratic, without exception, splits into two linear factors once complex numbers are available. The simplest cases are the most striking: $x^2 + 25 = x^2 - (5i)^2 = (x - 5i)(x + 5i)$ is a difference of squares in disguise, and $x^2 - 4x + 5$ , irreducible over the reals, factors as $(x - 2 - i)(x - 2 + i)$ .

Step back and look at what happened to the discriminant’s trichotomy. Over the reals it counted roots: two, one, or none. Over the complex numbers the count is always the same, exactly two roots counted with multiplicity, and the discriminant’s job shrinks to saying where they live. If $\Delta > 0$ they sit at two points of the real line; if $\Delta = 0$ they coincide at one real point. If $\Delta < 0$ they leave the line and stand as a conjugate pair, one above it and one below. Nothing is ever missing. This is the degree-two case of a much bigger theorem, the Fundamental Theorem of Algebra, which a later chapter will state for polynomials of every degree. That theorem reads: over the complex numbers, a degree- $n$ polynomial always has exactly $n$ roots with multiplicity. Quadratics are where that promise is first kept.

Worked example 5 Designing the root type

For which real values of $k$ does $x^2 + kx + 9 = 0$ have a conjugate pair of non-real roots?

Non-real roots happen exactly when the discriminant is negative. With $a = 1$ , $b = k$ , $c = 9$ ,

\Delta = k^2 - 36 < 0 \quad\Longleftrightarrow\quad k^2 < 36 \quad\Longleftrightarrow\quad -6 < k < 6,

solving the quadratic inequality as in Chapter 4. This is the same equation whose knife edge you found in the Quadratic Formula lesson: there, $k = \pm 6$ produced the single repeated root. Now the whole design space is mapped. For $-6 < k < 6$ the roots are the conjugate pair $-\tfrac{k}{2} \pm \tfrac{\sqrt{36 - k^2}}{2}\,i$ ; at $k = \pm 6$ they merge into the double root $\mp 3$ ; and for $|k| > 6$ they separate along the real line. As $k$ sweeps through, the two roots move continuously, meeting on the axis and then splitting off it. No value of $k$ leaves the equation rootless.

Common mistakes

Multiplying radicals before converting to $i$ -form. $\sqrt{-4}\cdot\sqrt{-9}$ is $(2i)(3i) = -6$ , not $\sqrt{36} = 6$ . The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ was proved for non-negative reals only. Rewrite every $\sqrt{-k}$ as $i\sqrt{k}$ first, then multiply.
Dividing only one term by $2a$ . $\dfrac{6 \pm 4i}{2}$ is $3 \pm 2i$ , never $3 \pm 4i$ or $6 \pm 2i$ . The denominator sits under the entire numerator, real part and imaginary part alike.
Answering “no solutions.” A negative discriminant means no real solutions. The equation has exactly two complex ones, so name the number system when you report the count.
Computing the conjugate product as $p^2 - q^2$ . $(p + qi)(p - qi) = p^2 - q^2 i^2 = p^2 + q^2$ : the $i^2 = -1$ flips the sign to a plus. The product of a non-real conjugate pair is always positive.
Using the conjugate theorem without real coefficients. $x^2 - 2ix - 1 = (x - i)^2$ has the double root $i$ , and $-i$ is not a root. The theorem requires every coefficient to be real; check that hypothesis before pairing roots.
Conjugating more than the imaginary sign. The conjugate of $-3 + 7i$ is $-3 - 7i$ . Only the sign of the imaginary part flips; the real part keeps its own sign.
Writing $\sqrt{-16} = \pm 4i$ . The symbol $\sqrt{-16}$ names one number, $4i$ , by the same convention that makes $\sqrt{9}$ mean $3$ . The second root of the equation comes from the $\pm$ that the quadratic formula already carries, not from the radical.

Key takeaways

For $k > 0$ the two square roots of $-k$ are $\pm i\sqrt{k}$ , and the symbol $\sqrt{-k}$ names $i\sqrt{k}$ . Convert to $i$ -form before doing any radical arithmetic.
When $\Delta = b^2 - 4ac < 0$ , the quadratic formula runs to completion and gives $x = -\dfrac{b}{2a} \pm \dfrac{\sqrt{4ac - b^2}}{2a}\,i$ , a conjugate pair $p \pm qi$ with $q \ne 0$ whose shared real part is the axis of symmetry.
Conjugate root theorem: if a quadratic has real coefficients and $z$ is a root, then $\overline{z}$ is a root, because conjugation passes through sums and products and fixes real numbers. Non-real roots never appear alone.
The root relations survive: sum $= 2p = -\dfrac{b}{a}$ and product $= p^2 + q^2 = \dfrac{c}{a} = |z|^2 > 0$ , so one non-real root rebuilds the monic quadratic as $x^2 - 2px + \left(p^2 + q^2\right)$ .
Over the complex numbers every quadratic factors as $a(x - z)(x - \overline{z})$ and has exactly two roots counted with multiplicity. The discriminant now only says where they live: on the real line, doubled on it, or off it as a mirror pair.

Complex Roots of Quadratics

Learning goals

Square roots of negative numbers, pinned down

The formula runs to completion

Non-real roots always arrive in conjugate pairs

Sum and product still tell the truth

Every quadratic factors completely now

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)