How to Solve Quadratics with Complex Numbers

A quadratic equation is an equation of degree 2 and is generally written as:

ax² + bx + c = 0, where a, b, and c are constants.

When solving these equations, the solutions can be real or complex numbers. Complex roots occur when the discriminant (b² −4ac) is negative. In such cases, complex numbers are used to find the roots.

A complex number contains a real part and an imaginary part. The imaginary unit is defined as: i = √-1 or i² = -1

Examples of complex numbers: 3 + 2i

Steps to Solve

Step 1: Ensure the quadratic equation is in standard form: ax² + bx + c = 0

Step 2: Calculate the Discriminant. The discriminant (Δ) is calculated as: Δ = b² − 4ac

Step 3: Determine the Nature of the Roots

• If Δ > 0, the roots are real and distinct.
• If Δ = 0, the roots are real and equal.
• If Δ < 0, the roots are complex conjugates.

Since we are focusing on complex roots, we consider Δ < 0.

Step 4: For complex roots, if the discriminant is negative, convert the square root into an imaginary number using iota(i).

Thus, \sqrt{\Delta} = \sqrt{-k} = i\sqrt{k}

Step 5: Substitute Δ = i√k​ into the quadratic formula:

x = \frac{-b \pm i\sqrt{k}}{2a}

This yields the complex roots:

x_1 = \frac{-b + i\sqrt{k}}{2a} and x_2 = \frac{-b - i\sqrt{k}}{2a}

Solved Example

Example 1: Solve the quadratic equation x² + 4x + 13 = 0.

Write in standard form: x² + 4x + 13 = 0

Calculate the discriminant: Δ = 4² − 4 ⋅ 1 ⋅ 13 = 16 − 52 = −36

Determine the nature of the roots: Since Δ < 0, the roots are complex.

Use the quadratic formula: x = [− 4 ± √(−36)]/(2 ⋅ 1) = (−4 ± 6i)/2

Simplify: x₁ = (−4 + 6i)/2 = −2 + 3i and x₂ = (−4 - 6i)/2 = −2 - 3i

Thus, the solutions are x₁ = −2 + 3i and x₂ = −2 − 3i.

Example 2: Solve x² + 6x + 13 = 0

Compare with ax² + bx + c = 0 so, a = 1 , b = 6 , c = 13

Using quadractic formula: x = (-6 ± √(36 - 52))/2

x = (-6 ± √(-16))/2

x = (-6 ± 4i)/2

x = -3 ± 2i

Thus, the solutions are :-3 + 2i and -3 - 2i

Example 3: Solve 2x² + 8x + 10 = 0

Here, a = 2, b = 8 ,c = 10

Using the quadratic formula:

x = (-8 ± √(64 - 80))/4

x = (-8 ± √(-16))/4

x = (-8 ± 4i)/4

x = -2 ± i

Thus, the solutions are : x = -2 + i and -2 - i

Example 4: Solve x² - 2x + 10 = 0

Here, a = 1 , b = -2, c = 10

Using the quadratic formula:

x = (2 ± √(4 - 40))/2

x = (2 ± √(-36))/2

x = (2 ± 6i)/2

x = 1 ± 3i

Thus, the solutions are : x = 1 + 3i and 1 - 3i

Practice Problems

1. Solve x² + 9 = 0

2. Solve x² + 2x + 5 = 0

3. Solve x² - 4x + 20 = 0

4. Solve 3x² + 6x + 15 = 0

5. Solve 2x² - 8x + 20 = 0

Related Article

• Complex Numbers
• Imaginary Numbers
• Quadratic Equation
• Quadratic Formula