Complex Numbers and Quadratic Solutions

Students use the imaginary unit and complex-number operations to represent and calculate nonreal solutions of quadratic equations.

Why Complex Numbers Are Needed

Some quadratic equations cannot be solved using only real numbers. For a quadratic equation ax² + bx + c = 0, the discriminant b² − 4ac indicates the kind of solutions it has. A positive discriminant gives two real solutions, zero gives one repeated real solution, and a negative discriminant gives two nonreal solutions. For example, x² + 4 = 0 leads to x² = −4. No real number has a square of −4 because the square of every nonzero real number is positive. On a graph, the parabola y = x² + 4 never crosses the x-axis, confirming that there are no real zeros. Complex numbers extend the real number system so that equations such as this one still have solutions.

A discriminant decision diagram appears beside the parabola y = x² + 4 staying above the x-axis.Source: Illustrated for this lesson

The Imaginary Unit

The imaginary unit, written i, is defined by i² = −1. Therefore, i can represent the square root of −1, and square roots of other negative numbers can be expressed using i. For example, x² = −4 has the solutions x = 2i and x = −2i because (2i)² = 4i² = −4 and (−2i)² = 4i² = −4. A complex number has the form a + bi, where a and b are real numbers. The number a is the real part, and b is the coefficient of the imaginary part. Powers of i repeat in a cycle: i, −1, −i, 1. For instance, i⁶ = i⁴ · i² = 1 · (−1) = −1.

A four-position powers-of-i cycle surrounds the standard form a + bi and a verification of the solutions to x² = −4.Source: Illustrated for this lesson

Simplifying Square Roots

To simplify the square root of a negative number, first factor out −1 and replace √−1 with i. Then simplify the remaining positive square root. For example, √−72 = √(−1 · 72) = i√72. Because 72 = 36 · 2, this becomes i√(36 · 2) = 6i√2. Thus, the principal square root of −72 is 6i√2. Be careful about the difference between simplifying a radical and solving an equation. The expression √−72 refers to one principal square root, but the equation x² = −72 has two solutions: x = ±6i√2. To check, squaring either value gives 36 · i² · 2 = 72(−1) = −72.

A worked radical chain simplifies √−72 to 6i√2 and contrasts it with both solutions of x² = −72.Source: Illustrated for this lesson

Operations with Complex Numbers

Complex numbers are added and subtracted by combining real parts with real parts and imaginary terms with imaginary terms. For example, (3 + 2i) + (5 − 7i) = 8 − 5i. To multiply, use the distributive property and replace i² with −1. Thus, (2 + 3i)(4 − i) = 8 − 2i + 12i − 3i² = 11 + 10i. Division often uses a conjugate. The conjugate of a + bi is a − bi, and their product is the real number a² + b². For example, 1/(2 + i) is multiplied by (2 − i)/(2 − i), giving (2 − i)/5. After any operation, write the result in standard form a + bi and simplify all powers of i.

Three worked examples show complex addition, multiplication, and division by a conjugate, each ending in standard form.Source: Illustrated for this lesson

Nonreal Quadratic Solutions

The quadratic formula solves equations with real coefficients even when the discriminant is negative. Consider x² − 4x + 13 = 0, where a = 1, b = −4, and c = 13. The discriminant is b² − 4ac = 16 − 52 = −36. Substitution into the formula gives x = (4 ± √−36)/2. Since √−36 = 6i, the solutions are x = (4 ± 6i)/2 = 2 ± 3i. These two answers, 2 + 3i and 2 − 3i, are complex conjugates. Nonreal solutions of a quadratic with real coefficients always occur as a conjugate pair. The graph y = x² − 4x + 13 has no x-intercepts, but the quadratic formula still identifies its two complex zeros.

The quadratic formula produces the conjugate pair 2 + 3i and 2 − 3i beside a parabola with no x-intercepts.Source: Illustrated for this lesson