Quadratic equations with complex roots

When the discriminant is negative

Take $ax^2 + bx + c = 0$ with real $a$, $b$, $c$ and $a \ne 0$. Its discriminant is $D = b^2 - 4ac$. In Class 10, a negative $D$ meant no real roots. Now we have $i$, and $\sqrt{D} = i\sqrt{-D}$. So the formula still works:

Take $x^2 - 4x + 13 = 0$. Here $D = 16 - 52 = -36$, so $\sqrt{D} = 6i$. The roots are

The parabola $y = x^2 - 4x + 13$ never meets the x-axis, so there is no real root. But the two complex roots are there, side by side on the Argand plane.

Start with $c$ small, then raise it. The two real roots move together, meet, and then leave the axis as a mirror pair.

Why the roots come in pairs

Let $p(z) = az^2 + bz + c$ with real $a$, $b$, $c$, and let $p(\alpha) = 0$. Take the conjugate of both sides. The rules for conjugates give

because $\bar a = a$, $\bar b = b$ and $\bar c = c$. So $\bar\alpha$ is a root too. The same argument works for a polynomial of any degree with real coefficients.

Sum and product of the roots

The usual rules still hold. With roots $\alpha$ and $\bar\alpha$:

• the sum is $\alpha + \bar\alpha = 2 \times (\text{real part of } \alpha) = -\dfrac{b}{a}$;
• the product is $\alpha\bar\alpha = |\alpha|^2 = \dfrac{c}{a}$.

So a quadratic with leading coefficient $1$ and one root $\alpha$ is

For example, if one root is $4 - i$, the equation is $x^2 - 8x + 17 = 0$.

A worked example

Let us solve $2x^2 + 3x + 4 = 0$.

Step 1. The discriminant. $D = 9 - 4 \times 2 \times 4 = 9 - 32 = -23$.

Step 2. Its square root. $\sqrt{-23} = i\sqrt{23}$.

Step 3. The roots. $x = \dfrac{-3 \pm i\sqrt{23}}{4}$, about $-0.75 \pm 1.199i$.

Step 4. Check. The sum is $-\dfrac{3}{2} = -\dfrac{b}{a}$, and the product is $\dfrac{9 + 23}{16} = 2 = \dfrac{c}{a}$.

The square root of a complex number

What are the square roots of $5 + 12i$? Look for $x + iy$, with $x$ and $y$ real, such that $(x + iy)^2 = 5 + 12i$. Expand and compare parts:

Take moduli too: $|x + iy|^2 = |5 + 12i|$, so $x^2 + y^2 = 13$.

Step 1. Add and subtract. $2x^2 = 18$ and $2y^2 = 8$, so $x = \pm 3$ and $y = \pm 2$.

Step 2. The signs. $2xy = 12$ is positive, so $x$ and $y$ have the same sign.

Step 3. The roots. $3 + 2i$ and $-3 - 2i$. Check: $(3 + 2i)^2 = 9 + 12i - 4 = 5 + 12i$.

In the same way, $-8 - 6i$ has modulus $10$. Then $x^2 = 1$ and $y^2 = 9$, with opposite signs, so its square roots are $\pm(1 - 3i)$.

Practise with new numbers

Try it in Python

The cmath module takes square roots of negative numbers. So one formula solves every quadratic. The program solves three and checks the sum and the product of the roots.

Common mistakes

• Saying there is no solution. A negative $D$ means no real root. There are still two complex roots.
• Losing the $i$. $\sqrt{-36}$ is $6i$, not $-6$.
• Dividing only one term. In $\dfrac{4 \pm 6i}{2}$, divide both $4$ and $6i$ by $2$.
• Ignoring the sign of $2xy$. For a square root, the sign of the imaginary part decides whether $x$ and $y$ match.

Remember these

• If $D < 0$, the roots are $\dfrac{-b \pm i\sqrt{-D}}{2a}$.
• With real coefficients, complex roots come in conjugate pairs.
• The sum of the roots is $-\dfrac{b}{a}$, and the product is $\dfrac{c}{a}$.
• For $\sqrt{a + ib}$, solve $x^2 - y^2 = a$, $2xy = b$ and $x^2 + y^2 = \sqrt{a^2 + b^2}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

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

2. Kiran says that $3 + 5i$ solves a quadratic whose coefficients are all real. Name its other root.

3. Find the square roots of $2i$.

4. Can a quadratic with real coefficients have the roots $2 + i$ and $3 - i$?

5. Find the discriminant of $x^2 + x + 1 = 0$, and the roots.