Square Root Calculator – Simplify Radicals
Enter any number — integer, decimal or fraction — and this square root calculator gives the decimal value, checks whether it is a perfect square, and shows the simplified radical step by step using prime factorization.
Simplify a Square Root, Step by Step
Square Root Formula
Where:
- \(x\) — the number under the radical sign, called the radicand
- \(a\) — the whole-number coefficient pulled out of the root
- \(b\) — the squarefree part that stays inside the radical (no perfect square greater than 1 divides it)
To find \(a\) and \(b\), write the prime factorization \(x = p_1^{e_1} \cdot p_2^{e_2} \cdots\). Each prime leaves the root with half its exponent and keeps the remainder inside: \(a = p_1^{\lfloor e_1/2 \rfloor} \cdot p_2^{\lfloor e_2/2 \rfloor} \cdots\) and \(b = p_1^{e_1 \bmod 2} \cdot p_2^{e_2 \bmod 2} \cdots\). For fractions, the same rule applies to the numerator and denominator separately, then the denominator is rationalized.
If \(b = 1\), no radical remains and \(\sqrt{x} = a\) is a whole number — a perfect square. If no perfect square greater than 1 divides \(x\), then \(a = 1\) and the radical is already in its simplest form.
Square Root – Calculation Example
Example: simplify \(\sqrt{72}\), the calculator's default value.
\[ 72 = 2^3 \cdot 3^2 \]
\[ \sqrt{72} = \sqrt{2^3 \cdot 3^2} = \sqrt{(2 \cdot 3)^2 \cdot 2} = 6\sqrt{2} \]
The prime 2 appears three times: one pair leaves the root as a factor of 2, and a single 2 stays inside. The pair of 3s leaves the root as one factor of 3, so the coefficient is \(2 \cdot 3 = 6\).
Result: \(\sqrt{72} = 6\sqrt{2} \approx 8.48528137423857\).
How to use the calculator
- Pick the Mode. Square root √x is the default. Choose n-th root ⁿ√x if you need a cube root, fourth root or any other order — an extra Root index (n) field appears for that.
- Enter the Number (x). Type an integer (72), a decimal (2.25), or a fraction (3/4). A leading minus sign is allowed: even roots of negative numbers come back as imaginary, like \(\sqrt{-16} = 4i\), while odd roots stay real, like \(\sqrt[3]{-27} = -3\).
- Read the results. Root value shows the decimal answer; Perfect square? tells you whether the root is a whole (or rational) number; Simplified radical form gives the exact answer such as \(6\sqrt{2}\); Steps: prime factorization shows the full simplification with every prime exponent; and the Perfect squares 1–30 table helps with hand calculations.
What does it mean to simplify a square root?
The square root of \(x\) is the non-negative number that, multiplied by itself, gives \(x\): \(\sqrt{x} \cdot \sqrt{x} = x\). For perfect squares such as 25 or 144, the root is a whole number — 5 and 12 respectively. For every other positive number, the root is irrational: its decimal never ends and never repeats, so a calculator can only show an approximation.
Simplifying a radical means rewriting that irrational root in its shortest exact form. The idea is to pull out every perfect square hiding inside the radicand. Since \(72 = 36 \cdot 2\), you get \(\sqrt{72} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}\). Nothing more can come out, because 2 has no square factor — the radical is now fully simplified. This exact form is preferred in algebra: it is precise, it makes further steps (like adding \(\sqrt{8} = 2\sqrt{2}\) to \(6\sqrt{2}\)) possible, and it reveals when two apparently different radicals are actually the same number.
The Perfect square? line tells you the nature of your answer. If it says yes, the root is rational and the radical disappears entirely. If it says no, the number lies between two consecutive perfect squares — for example, \(\sqrt{40}\) is between \(\sqrt{36} = 6\) and \(\sqrt{49} = 7\) — and the simplified radical is the only exact way to write it.
Fractions are handled exactly. The input is first reduced to lowest terms, then the denominator is rationalized: \(\sqrt{3/4} = \sqrt{3}/2\), not a rounded decimal. Decimals up to twelve places are converted to fractions internally, so the simplification stays exact.
Assumptions, limits and tips
- Principal root. For a positive \(x\), the calculator returns the positive (principal) root. Remember that \(x^2 = 72\) has two solutions: \(\pm 6\sqrt{2}\).
- Negative numbers. An even root of a negative number is imaginary: \(\sqrt{-16} = 4i\). In n-th root mode with an odd index, the answer stays real: \(\sqrt[3]{-27} = -3\).
- Size limits. After simplification, the numerator and denominator may each be at most 1012, with at most 13 significant digits entered — enough for any school, textbook or competition problem.
- Decimal answers are shown with up to 15 significant digits, so the last digit may differ from a value you round differently. Use the Rounding Calculator when a specific precision is required.
- Simplify by hand: divide the radicand by the largest perfect square factor you can spot from the 1–30 table, or check individual divisors with the Factor Calculator.
- The step-by-step breakdown relies on prime factorization — see the Prime Factorization Calculator for that part in isolation, or the Nth Root Calculator for roots of any order in full detail.
Frequently Asked Questions
How do I simplify a square root by hand without prime factorization?
Find the largest perfect square that divides the radicand and split the root. For \(\sqrt{200}\), notice \(200 = 100 \cdot 2\), so \(\sqrt{200} = \sqrt{100} \cdot \sqrt{2} = 10\sqrt{2}\). Test perfect squares from the 1–30 table (4, 9, 16, 25…) as divisors, starting from the largest. If only a small one works, like \(\sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}\), you may need a second pass — or just factorize to get it right the first time.
What is a perfect square, and how do I check if a number is one?
A perfect square is a number that equals some integer times itself: \(k \cdot k\). To check, take the square root and see whether the decimal part is zero — \(\sqrt{625} = 25\), so 625 is a perfect square. Prime factorization gives an exact test without rounding worries: every prime exponent must be even. For example \(625 = 5^4\) passes, while \(72 = 2^3 \cdot 3^2\) fails because the exponent of 2 is odd.
Why is \(6\sqrt{2}\) considered 'simpler' than \(\sqrt{72}\)?
Both are the same exact number, but \(6\sqrt{2}\) uses the smallest possible radicand, which makes radicals easier to compare and combine. You can add \(6\sqrt{2} + 3\sqrt{2}\) to get \(9\sqrt{2}\) directly, while \(\sqrt{72}\) and \(\sqrt{18}\) hide that they are like terms. Standardized tests and textbooks require answers in fully simplified radical form, meaning no perfect square greater than 1 divides the number under the root.
Can a square root be negative?
The symbol \(\sqrt{x}\) always means the principal (non-negative) root, so \(\sqrt{9} = 3\), never \(-3\). But an equation like \(x^2 = 9\) has two solutions, \(\pm 3\), and both matter in algebra and geometry problems. Separately, the square root of a negative number is not real: \(\sqrt{-9} = 3i\), where \(i\) is the imaginary unit with \(i^2 = -1\). Odd roots behave differently — \(\sqrt[3]{-27} = -3\) is a real number.
How do you simplify the square root of a fraction like \(\sqrt{8/50}\)?
Reduce the fraction first: \(8/50 = 4/25\). Then take the root of the numerator and denominator separately: \(\sqrt{4/25} = 2/5\). If the result has a radical in the denominator, such as \(\sqrt{3}/2\) from \(\sqrt{3/4}\), it is already rationalized; but \(\sqrt{1/2}\) becomes \(\sqrt{2}/2\), not \(1/\sqrt{2}\), because convention says no roots are allowed in the denominator.
What common mistakes should I avoid when simplifying radicals?
The biggest ones: assuming \(\sqrt{a+b} = \sqrt{a} + \sqrt{b}\) (false — check \(\sqrt{9+16} = 5\) but \(3+4 = 7\)), stopping at an incomplete simplification like \(\sqrt{72} = 2\sqrt{18}\) instead of \(6\sqrt{2}\), and adding unlike radicals, since \(\sqrt{2} + \sqrt{3}\) cannot combine. Also remember the root only distributes over multiplication and division, never over addition or subtraction inside the radicand.
How accurate is the decimal value, and why does \(\sqrt{2}\) never end?
The root of any positive number that is not a perfect square is irrational, so its decimal goes on forever without a repeating pattern — no calculator can write it exactly. \(\sqrt{2} \approx 1.41421356237310\) is shown with up to 15 significant digits here. That is why the simplified radical form \(6\sqrt{2}\) is preferred in exact work: it is completely precise, while any decimal is a rounded approximation. Use the rounding convention your instructor requires.