In this explainer, we will learn how to solve quadratic equations with complex coefficients using the quadratic formula.

From the fundamental theorem of algebra, we know that any quadratic will have two roots. For quadratic equations with real coefficients, the conjugate root theorem tells us that if it has any nonreal root, its roots will be a complex conjugate pair, whereas if it has real roots, it could either have two distinct real roots or a single repeated root. To distinguish between these three different cases, we use the discriminant.

### Discriminant

The discriminant of a quadratic equation is defined as . Often is used to denote the discriminant.

Using the discriminant, we identify the three different cases of quadratic equations as follows.

**Positive discriminant**: , two real roots**Zero discriminant**: , one repeated real root**Negative discriminant**: , complex conjugate roots

The graphs below depict each case.

In this explainer, we will relax the condition that the quadratic equation has real coefficients and explore what we can conclude about the roots of quadratic equations with complex coefficients.

Let us consider how to solve a general quadratic equation where . We start by dividing the equation by , which gives

We can now complete the square

Rearranging, we get

All of the steps we have taken up to this point are equally valid whether , , and are real or imaginary. The next step, where we take the square root, is something we might need to be careful with, since taking roots of a complex number returns multiple values. However, let us consider the case of a general complex number . Using de Moivreβs theorem, the two possible square roots of this number are and . Considering the second root , we have

Using the properties of exponential functions, we can express this as

However, courtesy of the Eulerβs identity, we know ; therefore,

Therefore, in the same way that we take both the positive and negative roots for real numbers, we can find a root of a complex number using de Moirveβs theorem, and then the second root will be its negative. Hence, returning to equation (1), we can take square roots and consider both the positive and negative values, which gives

Therefore,

Hence, even for quadratics with complex coefficients, we can use the quadratic formula, although we might need to appeal to de Moirveβs theorem to calculate the value of the square root of the discriminant for nonreal numbers.

In the first couple of examples, we will consider the special case where, in spite of the fact that we have some nonreal coefficients, the discriminant is still a real number.

### Example 1: Nonreal Quadratic Equations with Real Negative Discriminants

Solve .

### Answer

Using the quadratic formula we have

Hence, and are the solutions of the equation.

Notice that in the previous example, in spite of the fact that we had a negative discriminant, the two solutions are not a complex conjugate pair. Hence, the rule regarding a negative discriminant does not simply generalize to equations with complex coefficients.

### Example 2: Nonreal Quadratic Equations with Real Positive Discriminants

Solve .

### Answer

Using the quadratic formula we have

Hence, and are the solutions of the equation.

In the previous example, we had a positive discriminant; however, the two solutions are not real. Therefore, the rule regarding the positive discriminant does not simply generalize to equations with complex coefficients.

### Example 3: Nonreal Quadratic Equations with Real Discriminants

Solve .

### Answer

Applying the quadratic formula we have

Taking the 4 out of the square root, we have

Canceling the factor of 2 in the numerator and denominator, we have

Notice also that . Hence, is twice the complex number times its conjugate. Therefore, , and thus

Multiplying both the numerator and denominator by the complex conjugate of the denominator gives

Hence, and are the solutions of the equation.

Notice that, unlike the first two examples, the two roots in the previous examples do not lie on a vertical or horizontal line in the complex plane. The following figures represent the graph of the modulus of each equation plotted above the complex plane. The two points where the graph meets the complex plane are the roots of the equations.

### Example 4: Nonreal Quadratic Equations with Zero Discriminants

Solve .

### Answer

Using the quadratic formula we have

Hence, the equation has a single repeated root of .

The previous example demonstrates that it is possible to have repeated nonreal roots. Furthermore, we see that when this happened, the discriminant was zero. By simply considering the quadratic formula, we can see that the rule regarding a determinant of zero generalizes to the case where we have nonreal coefficients. This is in fact one of the key properties of the discriminant: that it is zero if, and only if, the equation has a repeated root.

In the following examples, we will consider the case where the discriminant is a general complex number. We will begin with a simple example where the coefficient of is zero, which of course is equivalent to simply finding the square roots of a complex number.

### Example 5: Solving Quadratic Equations by Taking the Square Root of Complex Numbers

Solve . Round your answers to three significant figures.

### Answer

Adding to both sides of the equation gives

We now divide by to isolate on the left-hand side of the equation:

By multiplying the numerator and denominator by the complex conjugate of the denominator, we can simplify the complex fraction as follows:

To find the square roots of this complex number, we need to calculate its modulus and argument. The modulus is given by . Since it lies in the fourth quadrant, its argument is given by

Applying de Moivreβs theorem, the two square roots are given by

Substituting in the value of and rounding our answers to three significant figures, we have and

### Example 6: Quadratic Equations with Complex Discriminants

Solve .

### Answer

Using the quadratic formula we have

To evaluate the square root of , we first need to calculate the modulus and argument. The modulus is given by . Since the complex number lies in the first quadrant, its argument is given by . Applying de Moivreβs theorem, the two square roots are given by

Substituting in the value of , we find and . Hence, the two square roots are given by and . Substituting these value back into (2), we have

Hence, the two roots of the equation are and .

We finish by considering one final example.

### Example 7: Polynomials with Nonreal Coefficients

Solve .

### Answer

We first notice that we have a quadratic in . Therefore, we start by making the substitution , which gives

We can solve this using the quadratic formula as follows:

Canceling the 2 from the numerator and denominator, we have

To evaluate the square root of , we need to first calculate its modulus and argument. The modulus is given by . Since this complex number lies in the second quadrant, its argument is given by

Using de Moivreβs theorem, its square roots are given by

Converting this back into algebraic form, we have that the two roots are and . Substituting these into (3), we get

Hence, the two roots are and . Finally, to find the possible values of , we need to take the square roots of each of these complex numbers, starting with . We can express this in exponential form as . Applying de Moivreβs theorem, the two square roots are given by which we can express in algebraic form as and . Similarly for , we can express this in exponential form as . Applying de Moivreβs theorem, the two square roots are given by , which we can express in algebraic form as and .

Therefore, the four roots of the equation are given by , , , and .

### Key Points

- We can solve quadratic equations with complex coefficients using the quadratic formula.
- If the discriminant is zero, the equation has one repeated root. If all the coefficients are real, the root will be real.
- A general quadratic with complex coefficients can have any combination of real and nonreal roots.
- If a quadratic has real coefficients, the discriminant is sufficient to completely classify the types of solution.