2.3: Complex Solutions | Algebra 2 Honors

How it works

Complex numbers extend real numbers by defining i squared to be -1. Negative radicands in the quadratic formula then give complex conjugate roots. When completing a square or taking a square root, include both signs unless a domain restriction says otherwise.

i^2=-1,\qquad\sqrt{-a}=i\sqrt{a}\quad(a>0)

Worked examples

Example 1

Solve $\displaystyle x^2+9=0$ .

$\displaystyle x^2=-9$ , giving $\displaystyle x=\pm3i$ .

Example 2

Solve $\displaystyle x^2-2x+5=0$ .

Complete the square: $\displaystyle (x-1)^2=-4$ . Hence $\displaystyle x=1\pm2i$ .

Problems

Problem 2.3.1

Solve $\displaystyle x^2+4=0$ .

Show worked solution

$\displaystyle x^2=-4$ , so $\displaystyle x=\pm2i$ .

Problem 2.3.2

Simplify $\displaystyle (3+2i)+(1-5i)$ .

Show worked solution

Combine real and imaginary parts: $\displaystyle 4-3i$ .

Problem 2.3.3

Multiply $\displaystyle (2+i)(2-i)$ .

Show worked solution

The product is $\displaystyle 4-i^2=5$ . Complex conjugates have a real product.

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 2.3.1

Problem 2.3.2

Problem 2.3.3

Need help with this topic?