Square Root Calculator

Calculate square roots for positive and negative numbers. For negative values, the result is shown as an imaginary number with i.

Square roots, including the negative case

Every positive number has two square roots, one positive and one negative, and the radical symbol conventionally denotes only the positive one. Negative inputs have no real root at all, which is where imaginary numbers enter - and they are far more practical than the name suggests.

Formula

For n >= 0: sqrt(n) = x where x^2 = n | For n < 0: sqrt(n) = i x sqrt(|n|)

The principal square root is always the non-negative one. sqrt(25) is 5, while the equation x^2 = 25 has both 5 and -5 as solutions.

Two roots, one symbol

This distinction causes more confusion than any other part of the topic. sqrt(9) = 3, full stop. But the equation x^2 = 9 has solutions x = 3 and x = -3, written x = +/-3.

The convention exists so that the square root can be a function - one input, one output. Without it, sqrt would be ambiguous and could not be composed or graphed.

The practical consequence: when you take the square root of both sides while solving an equation, you must introduce the +/- yourself. Forgetting it loses half the solutions, and in physics problems the discarded root is often the physically meaningful one.

Negative inputs and imaginary numbers

No real number squares to a negative, because both positive and negative numbers square to positive. Mathematics resolves this by defining i such that i^2 = -1.

Then sqrt(-4) = 2i, sqrt(-1) = i, and sqrt(-18) = 3i x sqrt(2), or about 4.243i. Everything else works as expected: the magnitude comes from the absolute value, and i carries the sign information.

One rule breaks: sqrt(a) x sqrt(b) = sqrt(ab) only holds when at least one of a and b is non-negative. Applying it to two negatives produces the classic false result that sqrt(-1) x sqrt(-1) = sqrt(1) = 1, when it is actually -1.

Despite the name, imaginary numbers describe measurable things. Alternating current impedance, quantum wavefunctions, signal processing, and control system stability all depend on them.

Estimating roots without a calculator

Bracket the number between two perfect squares, then interpolate. For sqrt(50): 49 is 7^2 and 64 is 8^2, so the answer is just over 7. Linear interpolation gives 7 + 1/15 = 7.07, and the true value is 7.0711.

For a better estimate, use one step of the Babylonian method: guess g, then improve it with (g + n/g) / 2. Starting from 7 for sqrt(50): (7 + 50/7) / 2 = 7.0714, accurate to four decimal places after a single iteration.

This method predates Newton by well over a millennium and converges quadratically, roughly doubling the number of correct digits each pass.

Worked example

A square plot has an area of 340 m^2. What is its side length?

sqrt(340) lies between sqrt(324) = 18 and sqrt(361) = 19. Babylonian step from 18: (18 + 340/18) / 2 = 18.444. Second step: (18.444 + 340/18.444) / 2 = 18.4391. The side is 18.44 m.

The negative root, -18.44, is mathematically valid but physically meaningless here - a length cannot be negative. This is the standard reason for discarding one root in applied problems, and it must be justified rather than assumed.

Perfect squares worth recognising

nn^2nn^2
1112117289
1214418324
1316919361
1419620400
1522525625
1625630900

Knowing squares up to 25 makes bracketing estimates fast and reliable.

Frequently Asked Questions

Because it is in real mode. The answer is 2i, which requires complex number support. Most scientific calculators have a complex or a+bi mode that will return it.

The radical symbol denotes the principal, non-negative root, so sqrt(25) = 5. The equation x^2 = 25 is a different statement and has both 5 and -5 as solutions.

Zero, and it is the only number with exactly one square root. Every positive number has two, and negatives have two imaginary ones.

Extensively. Electrical engineers use them for AC impedance, signal processing uses them for the Fourier transform, and quantum mechanics is formulated in complex Hilbert space. The name is a historical accident from when they were regarded as suspect.

Because assuming it equals a fraction in lowest terms leads to a contradiction: both numerator and denominator turn out to be even, so the fraction was not in lowest terms. This proof, attributed to the Pythagoreans, is one of the oldest in mathematics.

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References

Last reviewed: 2026-08-07. This page is informational. For legal, medical, tax, or financial decisions, confirm the result with a qualified professional.