Explainer: Solving Quadratic Equation with Complex Roots

In this explainer, we will learn how to solve quadratic equations whose roots are complex numbers.

The introduction of complex numbers opens up the possibility of finding solutions to equations which we were formerly unable to solve. For example, formerly, when we came across an equation that required us to take the square root of a negative number, we were unable to solve it and rightly concluded there were no real solutions. However, using complex numbers, we can gain additional insight by exploring the complex solutions to these equations. We begin by looking at an example of an equation we are unable to solve if we restrict ourselves to dealing solely with real numbers.

Example 1: Solving Equations with Complex Numbers

Solve the equation 5𝑥+1=319.

Answer

We start by gathering our like terms: 5𝑥=320.

Dividing both sides by 5, we isolate 𝑥 on the left of the equation as follows: 𝑥=3205=64.

Taking the square root of both sides remembering that we can take both the positive and the negative roots, we get 𝑥=±64.

Using the property of complex numbers that, for a positive number 𝑎, 𝑎=𝑖𝑎, we can rewrite this as 𝑥=±𝑖64=±8𝑖.

As seen above, the methods we apply to solving equations with real solutions can often be directly applied to equation with nonreal solutions. Hence, when dealing with quadratic equations, methods such as factoring and completing the square can equally be applied to equations with complex solutions. In particular, we can also use the quadratic formula.

Quadratic Formula

For a quadratic equation 𝑎𝑥+𝑏𝑥+𝑐=0 with 𝑎0, the roots are given by 𝑥=𝑏±𝑏4𝑎𝑐2𝑎.

Using the quadratic formula, we are able to solve any quadratic equation. When using the quadratic formula, we come across three distinct cases. To distinguish between them, we introduced the idea of the discriminant.

Discriminant

The discriminant of a quadratic equation 𝑎𝑥+𝑏𝑥+𝑐=0 is defined as 𝑏4𝑎𝑐. Often Δ is used to denote the discriminant.

Using the discriminant, we identify the three different case of quadratic equations as follows:

  1. positive discriminant: 𝑏4𝑎𝑐>0, two real roots;
  2. zero discriminant: 𝑏4𝑎𝑐=0, one repeated real root;
  3. negative discriminant: 𝑏4𝑎𝑐<0, no real roots.

The graphs below depict each case.

We will primarily focus on case number 3, in which there are no real solutions. The introduction of complex numbers enables us to restate this as the case were we have complex roots. In this explainer, we will explore this case and the properties of these roots.

Example 2: Solving Quadratic Equations with Complex Roots

Solve the quadratic equation 𝑥4𝑥+8=0.

Answer

Using the quadratic formula, substituting in 𝑎=1, 𝑏=4, and 𝑐=8, we have 𝑥=𝑏±𝑏4𝑎𝑐2𝑎=(4)±(4)4×1×82×1.

Simplifying, we get 𝑥=2±121632=2±1216.

Using the property of complex numbers that, for a positive number 𝑎, 𝑎=𝑖𝑎, we can rewrite this as 𝑥=2±12𝑖16=2±12×4𝑖=2±2𝑖.

The fact that the roots of this equation are a complex conjugate pair is no accident. In fact, it is true for any quadratic equation with real coefficients which has nonreal solutions as stated in the following theorem.

Conjugate Root Theorem for Quadratic Equations

The nonreal roots of a quadratic equation 𝑎𝑥+𝑏𝑥+𝑐=0 with real coefficients occur in complex conjugate pairs. Hence, if 𝑧=𝑚+𝑛𝑖 (where 𝑛0) is a root of a quadratic equation with real coefficients, then 𝑧=𝑚𝑛𝑖 is also a root.

Using the properties of the complex conjugate, we will provide a simple proof of this theorem. For a quadratic function 𝑓(𝑥)=𝑎𝑥+𝑏𝑥+𝑐, let 𝜂 be a complex root. Now, we consider 𝑓(𝜂)=𝑎(𝜂)+𝑏𝜂+𝑐.

From the properties of the complex conjugate, we know that for any two complex numbers (𝑧𝑧)=𝑧𝑧; hence, (𝜂)=(𝜂). Therefore, we have 𝑓(𝜂)=𝑎𝜂+𝑏𝜂+𝑐.

Given that any real number is equal to its complex conjugate and that 𝑎, 𝑏, and 𝑐 are real numbers, we can rewrite this as 𝑓(𝜂)=𝑎𝜂+(𝑏𝜂)+𝑐.

We also know from the properties of the complex conjugate that for any two complex numbers (𝑧±𝑧)=𝑧±𝑧; hence, 𝑓(𝜂)=𝑎𝜂+𝑏𝜂+𝑐=𝑎𝜂+𝑏𝜂+𝑐.

However, since 𝜂 is a root, we know that 𝑎𝜂+𝑏𝜂+𝑐=0. Therefore, 𝑓(𝜂)=0=0. Hence, we have shown that 𝜂 is also a root of 𝑓. Both this theorem and its proof can be generalized to any polynomial with real coefficients. In the remainder of this explainer, we will apply this theorem to a number of examples.

Example 3: Complex Roots of Quadratic Equations

The complex numbers 𝑎+𝑏𝑖 and 𝑐+𝑑𝑖, where 𝑎, 𝑏, 𝑐, and 𝑑 are real numbers, are the roots of a quadratic equation with real coefficients. Given that 𝑏0, what conditions, if any, must 𝑎, 𝑏, 𝑐, and 𝑑 satisfy?

Answer

Since 𝑏0, we know that 𝑎+𝑏𝑖 is a nonreal solution to a quadratic equation with real coefficients. Therefore, its complex conjugate will also be a root (curtsey of the conjugate root theorem). Furthermore, since a quadratic equation only has two roots, 𝑐+𝑑𝑖 must be the conjugate of 𝑎+𝑏𝑖. Hence, 𝑐+𝑑𝑖=(𝑎+𝑏𝑖)=𝑎𝑏𝑖.

Therefore, 𝑎=𝑐 and 𝑑=𝑏.

Example 4: Conditions on the Roots of Quadratic Equations

If the discriminant of a quadratic equation with real coefficients is negative, will its complex roots be a conjugate pair?

Answer

Given that the discriminant is negative, we know the equation has nonreal roots. Furthermore, since it has real coefficients, we know these roots will be a complex conjugate pair. Hence, the answer is yes.

We can also use our knowledge of the roots of quadratic equations with real coefficients to reconstruct an equation given one of its nonreal roots as the next example will demonstrate.

Example 5: Reconstructing a Quadratic Equation from a Nonreal Root

Find the quadratic equation with real coefficients which has 5+𝑖 as one of its roots.

Answer

Since the quadratic equation has real coefficients, its roots will be a complex conjugate pair. Hence, its roots will be 5+𝑖 and 5𝑖. Therefore, we can write the equation as (𝑥(5+𝑖))(𝑥(5𝑖))=0.

Using FOIL, or another similar method, we can expand the brackets as follows: 𝑥(5+𝑖)𝑥(5𝑖)𝑥+(5+𝑖)(5𝑖)=0.

Similarly, we can expand (5+𝑖)(5𝑖) and simplify: 𝑥10𝑥+255𝑖+5𝑖𝑖=0.

Using 𝑖=1, we have 𝑥10𝑥+26=0.

More generally, if we have a quadratic equation with roots 𝛼 and 𝛽, we can write the equation as (𝑧𝛼)(𝑧𝛽)=0.

Expanding the brackets, we get 𝑧(𝛼+𝛽)𝑥+𝛼𝛽=0.

This is true for any quadratic equation. However, if 𝛼 and 𝛽 are complex conjugates, then we have 𝑧(𝛼+𝛼)𝑧+𝛼𝛼=0.

Recall that, for any complex number 𝛼=𝑎+𝑏𝑖, 𝛼+𝛼=2𝑎 and 𝛼𝛼=𝑎+𝑏. Hence, if we know that 𝛼=𝑎+𝑏𝑖 is the root of a quadratic equation with real coefficients, we can conclude that the equation will be 𝑥2𝑎𝑥+𝑎+𝑏=0.

By applying this knowledge, it is possible to significantly simplify the calculation presented in the previous example.

Key Points

  1. Using the same methods developed to solve quadratic equations with real roots, we can solve quadratic equations with complex roots.
  2. The complex roots of quadratic equations with real coefficients occur in complex conjugate pairs. Hence, if 𝑧 is nonreal and a root of a quadratic equation with real coefficients, we know that 𝑧 is also a root.
  3. Given a single complex root to a quadratic equation with real coefficients, we can reconstruct the original equation. In particular, if one of the roots is 𝑎+𝑏𝑖, the quadratic equation will be 𝑥2𝑎𝑥+𝑎+𝑏=0.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.