Square root of negative number explained simply

When you try to find a square root of a negative number using a real-number calculator, you get an error message. That error is not a sign that you have done something wrong. In the real-number system, no number multiplied by itself ever produces a negative result. A positive times a positive is positive; a negative times a negative is also positive. There is no real number whose square is −1, −4, or −16. That is the entire reason the imaginary unit i was defined.

The problem is real, but the solution is not a real number. If you are solving an equation like x² = −9, you need a number that works. That number exists in the complex number system, and it is built from i = √(−1).

A square root of a given number is a value that, when squared, returns that number. For any real number, its square is zero or positive. The radicand in √(−4) is negative. No real number can be squared to give a negative result, so √(−4) has no real square root. This is not a limitation of the calculator; it is a fundamental property of real numbers. The radical symbol √ applied to a negative radicand produces an imaginary number, not a real one. That is why every spreadsheet SQRT function returns an error when the argument is negative.

The imaginary unit i is defined as i = √(−1). More precisely, i is a number that satisfies i² = −1. OpenStax Intermediate Algebra 2e, section 8.8, states this as the foundation of the complex number system. Every square root of a negative number can be expressed in terms of i. For example, √(−16) = √(16 × −1) = √16 × √(−1) = 4i. The result is an imaginary number. When combined with a real number, it forms a complex number in the standard form a + bi, where a is the real part and b is the coefficient of the imaginary part.

Powers of i Cycle

The powers of i follow a repeating pattern: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. After i⁴, the cycle repeats. This pattern is essential for simplifying expressions with higher powers of i. OpenStax Intermediate Algebra 2e, section 8.8, gives this cycle as a standard reference.

The rule for simplifying the square root of a negative number is: for any positive real number a, √(−a) = i√a. OpenStax Intermediate Algebra 2e, section 8.8, gives this formula as the correct method. Here are worked examples.

Example 1: √(−25)

Write the radicand as a product of −1 and a positive number: √(−25) = √(25 × −1) = √25 × √(−1) = 5i. The result is an imaginary number with real part 0 and imaginary part 5.

Example 2: √(−50)

First factor out −1: √(−50) = √(50 × −1) = √50 × √(−1). Simplify √50 to its simplest radical form: √50 = √(25 × 2) = 5√2. Then multiply by √(−1) = i: √(−50) = 5√2 i. The imaginary part here is 5√2.

Example 3: √(−72)

√(−72) = √(72 × −1) = √72 × i. Simplify √72: √72 = √(36 × 2) = 6√2. So √(−72) = 6√2 i.

Example 4: √(−1)

This is the defining case: √(−1) = i. This is not an approximation; it is a definition used in all complex number arithmetic.

A common failure occurs when applying the product rule √(a × b) = √a × √b to two negative numbers. This rule is valid only when a and b are non-negative. For example, √(−4 × −9) should not be written as √(−4) × √(−9). Doing that gives (2i)(3i) = 6i² = −6. But −4 × −9 = 36, and √36 = 6. The product rule fails because the radicands are negative. The correct approach is to multiply the numbers first: √(−4 × −9) = √36 = 6. This trap is one of the confusion pairs listed in research: the product rule holds only for non-negative a and b; for negatives, it leads to contradictions like 1 = √(−1 × −1) ≠ i × i = −1.

Square roots of negative numbers appear naturally when solving quadratic equations with a negative discriminant. The discriminant of a quadratic ax² + bx + c = 0 is Δ = b² − 4ac. When Δ is negative, the quadratic has no real solutions, but it has two complex solutions. The quadratic formula x = [−b ± √(b² − 4ac)] / (2a) requires the square root of the discriminant. If Δ is negative, you take the square root of a negative number, which introduces i. For example, solving x² + 4x + 5 = 0 gives Δ = 16 − 20 = −4. Then x = [−4 ± √(−4)] / 2 = [−4 ± 2i] / 2 = −2 ± i. The solutions are complex numbers of the form a + bi.

This is covered in OpenStax Intermediate Algebra 2e, section 8.8, under the complex number system. The same section defines the conjugate a − bi and the product of conjugates (a + bi)(a − bi) = a² + b², which is used to simplify division of complex numbers.