Virginia SOL Math / Algebra II / Equations & Inequalities

A2.EI.2

Quadratics Over the Complex Numbers

What you'll learn

  • Solve quadratic equations by factoring, roots, or the quadratic formula
  • Use the discriminant to describe the nature of the roots
  • Find complex solutions when the discriminant is negative

Words to know

Quadratic equation —
an equation of the form ax^2 + bx + c = 0
Quadratic formula —
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
Discriminant —
the value b^2 - 4ac
Complex root —
a solution of the form a + bi
Double root —
a repeated solution

Why it matters Over the complex numbers every quadratic has two solutions, so we can always describe where a parabola meets any line.

Worked examples

Model ItI do

Solve x^2 - 2x + 5 = 0.

Here a = 1, b = -2, c = 5. The discriminant is (-2)^2 - 4(1)(5) = 4 - 20 = -16, which is negative, so the roots are complex. The quadratic formula gives x = (2 +/- sqrt(-16)) / 2 = 1 +/- 2i. The parabola never crosses the x-axis.

Try It TogetherWe do

Solve x^2 - 5x + 6 = 0.

Factor: (x - 2)(x - 3) = 0. So x = 2 or x = 3. The discriminant is 25 - 24 = 1 > 0, giving two real roots.

Now YouYou do

Solve x^2 + 9 = 0.

Isolate x^2, then take the square root of a negative. Check your answer in the back of this chapter.

Practice

Mastered 0 of 11

Warm-up

Ease in with the basics.

  1. 1.Solve x^2 = 36.

  2. 2.Solve x^2 + 16 = 0.

  3. 3.Solve x^2 - 7x + 12 = 0.

On Level

The core Quick Check from the book.

  1. 1.Solve x^2 = 16.

  2. 2.Solve x^2 + 25 = 0.

  3. 3.Solve x^2 - 4x + 3 = 0.

  4. 4.Find the discriminant of x^2 + x + 1 = 0.

  5. 5.How many real roots does x^2 + 4 = 0 have?

Challenge

Stretch your thinking.

  1. 1.Solve x^2 - 4x + 13 = 0.

  2. 2.Solve 2x^2 + 3x - 2 = 0.

  3. 3.Describe the roots of x^2 - 6x + 9 = 0 using the discriminant.