Square Root Calculator
Square roots, cube roots and any nth root up to the twentieth — with the answer in simplified radical form as well as in decimal. √72 is 6√2, and that is the form most homework actually asks for.
Take a root
The number under the radical. Negatives are allowed — an odd root of one is real, an even root is imaginary.
2 for a square root, 3 for a cube root, and any whole number up to 20 for the rest.
√72 = 8.485281374. In simplified radical form that is 6√2, because 72 = 36 × 2.
- Simplified radical form
- 6√2
- Square root
- 8.485281374
- Cube root
- 4.160167646
- Perfect square
- No
- Perfect cube
- No
One definition, two answers, and a factorisation
An nth root undoes an nth power. That single sentence fixes everything on this page, including the parts that look like exceptions.
ⁿ√x = r where rⁿ = x • √(a²b) = a√b • √(−x) = i√x- n
- The degree — 2 for a square root, 3 for a cube root
- r
- The principal root: the non-negative one when there are two
- a
- The largest whole number whose square divides x — what comes out
- b
- What is left underneath, with no square factor in it
How the radical is simplified
The engine behind this page divides the number by 4, then 9, then 16, and so on, exhausting each candidate before moving to the next. Every time a perfect square divides cleanly, its root moves outside the radical and the quotient stays in. Because each factor is used up before the next is tried, a composite candidate can never claim a power its own prime factors have already removed — by the time the loop reaches 4, every factor of 2 is gone.
The same loop works for any degree; it just divides by cubes for a cube root and fifth powers for a fifth root. That is why ∛54 comes back as 3∛2 — 54 = 27 × 2 and 27 is 3³ — rather than as a bare decimal.
Negatives, and where most implementations break
An even root of a negative number has no real value, so the tool reports the magnitude and marks it imaginary: √(−4) is 2i. An odd root of a negative number is real, because an odd power keeps its sign — ∛(−27) is −3. The catch is that the usual way to compute an nth root, x1/n, is undefined for a negative base in almost every programming language: in JavaScript, (-27) ** (1/3) evaluates to NaN while Math.cbrt(-27) gives −3. Every odd root here is taken on the magnitude and signed afterwards, so it never meets that hole.
Precision
Square roots use the hardware square root, which IEEE 754 requires to be correctly rounded — so a perfect square comes back exact. Cube roots use a dedicated cube-root routine, and degrees above three fall back to a general power, which can leave noise in the last bit or two: 321/5 can evaluate to 2.0000000000000004. Results are rounded to twelve significant figures, which is inside that noise and finer than anything the number will be used for.
√72, carried all the way through
Start with the prime factorisation: 72 = 2 × 2 × 2 × 3 × 3, or 2³ × 3². A square root takes out one of every pair, so the pair of 2s yields a 2, the pair of 3s yields a 3, and the third 2 has no partner and stays put. Outside the radical: 2 × 3 = 6. Inside: 2. So √72 = 6√2.
Check it two ways. Squaring the answer, 6² × 2 = 36 × 2 = 72 ✓. Numerically, √2 = 1.414213562, so 6√2 comes to 8.485281374, which is what the calculator shows at the top of this page. The two forms are the same number; only one of them is exact.
Change the degree and 72 factorises differently. For a cube root, the pull-out is 2³ = 8, leaving 9 — so ∛72 = 2∛9 ≈ 4.160167646. For a fifth root, no fifth power divides 72 at all, so nothing comes out and the answer is the decimal 2.352158045 alone. Not every number has a tidy radical form, and the tool says so rather than inventing one.
The proof that √2 is irrational
8.485281374 is not √72; it is √72 rounded. The decimal never terminates and never repeats, and there is a short proof of why that is not a limitation of the calculator.
Suppose √2 could be written as a fraction p/q in lowest terms. Square both sides: 2 = p²/q², so p² = 2q². That makes p² even, and an odd number squared is always odd, so p must be even. Write p = 2k. Substituting back, 4k² = 2q², so q² = 2k² — and by the same argument q is even too. But p and q were in lowest terms, so they cannot share a factor of 2. The supposition collapses, so no such fraction exists.
Every square root of a non-square integer is irrational for the same reason, which is exactly why simplified radical form matters. 6√2 is the whole number; any decimal you write down is a truncation of it.
What this calculator assumes
- The principal root. Both 3 and −3 square to 9, and √9 means 3. If you are solving x² = 9 rather than evaluating √9, remember the ± yourself — it belongs to the equation, not to the symbol.
- Radical form needs a whole number. Simplification is attempted only on integers, and only up to a magnitude of a billion. √2.5 has no tidy radical form, and nobody factors √999999937 by hand.
- Imaginary results are a magnitude and a flag. √(−4) is reported as 2i, but this is not a complex-number calculator: it will not multiply two imaginary numbers together or find the other roots that lie off the real line.
- Every nth root has n roots. The tool returns the one real principal root. A fourth root has four complex fourth roots, evenly spaced around a circle; three of them are out of scope here.
- Twelve significant figures. Results are rounded once, at the end, from full double-precision arithmetic. Degrees above three carry a little last-bit noise before that rounding; square roots do not.
Square root FAQ
What is simplified radical form, and why does my teacher want it?
Simplified radical form means no perfect square is left under the radical sign: √72 becomes 6√2, because 72 contains the perfect square 36. Teachers want it for two reasons. It is exact — 6√2 is the number, while 8.485281374 is a truncation of it — and it is the only form you can do further algebra in. 6√2 plus 5√2 is 11√2 in one step; the decimals would have to be added and then re-recognised. Every algebra course marks the decimal as an incomplete answer for that reason.
How do I simplify a square root by hand?
Break the number into prime factors and pull out every pair. For 72: 72 = 2 × 2 × 2 × 3 × 3. There is a pair of 2s and a pair of 3s, so a 2 and a 3 come out — 2 × 3 = 6 — and the leftover 2 stays underneath. That gives 6√2. The shortcut version is to find the largest perfect square that divides the number: 36 divides 72, 72 ÷ 36 = 2, and √36 = 6, so √72 = 6√2. Both routes reach the same place; the prime factorisation is the one that never leaves a factor behind.
Why can you not take the square root of a negative number?
Because squaring destroys the sign. A positive number squared is positive, and a negative number squared is also positive, so nothing you square gives a negative result — there is no real number whose square is −4. Mathematicians defined one anyway: i, the number whose square is −1. Then √(−4) = √4 × √(−1) = 2i. That is what this calculator returns, and it is a real answer to the question, not an error. It is only in the real numbers that it does not exist.
Odd roots of negatives are fine though — why?
Because an odd power keeps the sign. (−3) × (−3) × (−3) = −27, since two negatives make a positive and the third makes it negative again. So ∛(−27) is exactly −3, a perfectly ordinary real number. This trips up a surprising amount of software, because the common implementation of an nth root is value^(1/n), and in most languages a negative base with a fractional exponent returns NaN — in JavaScript, (−27) ** (1/3) is NaN while Math.cbrt(−27) is −3. This tool takes every odd root of the magnitude and puts the sign back afterwards.
Is √2 really irrational, and how do we know?
Yes, and the proof is one of the oldest in mathematics — it is barely five lines. Suppose √2 were a fraction p/q already reduced to lowest terms. Squaring gives p² = 2q², so p² is even, and only an even number squares to an even number, so p is even. Write p = 2k. Then 4k² = 2q², so q² = 2k², so q is even too. But p and q were in lowest terms and cannot both be even. The assumption breaks, so no such fraction exists. The tradition that this discovery embarrassed the Pythagoreans is probably a later story, but the argument itself is genuinely ancient and appears in Aristotle as an example of proof by contradiction.
Why is √9 equal to 3 and not ±3?
Because the √ symbol is defined to mean the principal — non-negative — root, so that it names one number rather than two. Both 3 and −3 square to 9, but √9 is 3 by definition. The ± appears when you solve an equation rather than evaluate a symbol: x² = 9 has the two solutions x = 3 and x = −3, and it is the act of taking the root of both sides that introduces the ±. This calculator evaluates the symbol, so it returns the principal root.
Why does a calculator show 2.0000000000000004 for the fifth root of 32?
Because most nth roots are computed as exp(ln(x) ÷ n), and both the logarithm and the exponential are rounded to the nearest representable binary number along the way. The error is about one part in ten thousand million million, and it lands in the last digit. Square roots do not suffer from it: IEEE 754 requires square root to be correctly rounded, so √1024 is exactly 32 in any conforming implementation. This tool rounds every result to twelve significant figures, which is well inside the noise floor of the general nth-root path and well outside anything you would use the answer for.
Sources and review notes
- NIST Digital Library of Mathematical Functions §4.2 — the definitions of powers and roots, including the principal-value convention this tool follows
- IEEE 754-2019, Standard for Floating-Point Arithmetic — square root is one of the operations required to be correctly rounded, which is why a perfect square comes back exact here
The arithmetic on this page is definitional and has nothing behind it to go stale. What can go wrong is the implementation, so the engine is a pure function under unit test — including cases built specifically around the two traps above: the odd root of a negative number, and the last-bit noise in a general nth root.