The square root of a number \(a\) is a number whose square is \(a\), and the arithmetic square root refers only to its non-negative value: \(\sqrt{25} = 5\). You can only take the root of non-negative numbers, but the properties of the root turn cumbersome expressions into short ones. This page covers the definition and domain, all properties of the square root with formulas, a table of square roots from 1 to 20, methods for extracting roots and factoring out multipliers, solved examples, common mistakes, and an interactive trainer.
What is the square root of a number
The square root of a number \(a\) is a number whose square is \(a\). Extracting a root and squaring are inverse operations: from \(7^2 = 49\), it directly follows that the square root of 49 is 7.
However, a positive number has exactly two square roots, differing only by their sign: \(5^2 = 25\) and \((-5)^2 = 25\), meaning the square roots of 25 are both \(5\) and \(-5\). To ensure the notation \(\sqrt{25}\) denotes a single, unique number, it's agreed that the \(\sqrt{\phantom{a}}\) symbol represents only the non-negative root. This is called the arithmetic square root.
The notation has its own names: the \(\sqrt{\phantom{a}}\) symbol is the radical, and the number or expression beneath it is the radicand.
Arithmetic square root: definition and domain
The arithmetic square root of a non-negative number \(a\) is the non-negative number whose square is \(a\). It is denoted by \(\sqrt{a}\) and read as 'the square root of a'.
The definition contains two conditions, both equally important:
- the radicand must be non-negative: \(a \ge 0\);
- the value of the root must be non-negative: \(\sqrt{a} \ge 0\).
From this, we immediately get: \(\sqrt{25} = 5\), not \(-5\); \(\sqrt{0} = 0\); and the notation \(\sqrt{-9}\) has no meaning in real numbers – there is no number whose square is negative.
The first condition defines the domain of an expression with a root. For example, \(\sqrt{x-7}\) is meaningful when \(x - 7 \ge 0\), which means \(x \ge 7\): any smaller value of \(x\) would place a negative number under the root.
The definition also implies the equality \((\sqrt{a})^2 = a\) for \(a \ge 0\): squaring the root returns the original number. This holds true even when we cannot calculate the root precisely: \((\sqrt{31})^2 = 31\).
Table of square roots: squares of numbers from 1 to 20
A root is extracted exactly when the radicand is a perfect square. Therefore, the table of squares also works as a table of square roots: it can be read in both directions - \(14^2 = 196\), so \(\sqrt{196} = 14\).
The table is helpful not only for integers:
- decimal fractions: \(\sqrt{0{,}64} = 0{,}8\), because \(0{,}8^2 = 0{,}64\);
- common fractions: \(\sqrt{\dfrac{16}{81}} = \dfrac{4}{9}\) — the root is extracted separately for the numerator and the denominator;
- round numbers: \(\sqrt{2500} = 50\), since \(2500 = 25 \cdot 100\), and the root of each factor is known.
Properties of a square root
The properties of a square root allow calculations without extracting each root individually, and simplify the answer.
The root of a product is the product of the roots: \(\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}\) for \(a \ge 0\) and \(b \ge 0\). For example, \(\sqrt{9 \cdot 16} = 3 \cdot 4 = 12\) - the same result as \(\sqrt{144}\).
The root of a quotient is the quotient of the roots: \(\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}\) for \(a \ge 0\) and \(b > 0\). For example, \(\sqrt{\dfrac{36}{49}} = \dfrac{6}{7}\). The condition \(b > 0\) is strict: division by zero is not allowed.
The root of a square is the absolute value: \(\sqrt{a^2} = |a|\), and this is true for any number \(a\). Here lies the main subtlety of the entire topic. If \(a \ge 0\), nothing changes: \(\sqrt{7^2} = 7\). But if \(a\) is negative, the sign 'straightens out': \(\sqrt{(-6)^2} = \sqrt{36} = 6\), not \(-6\) – because an arithmetic root cannot be negative.
The square of a root returns the radicand: \((\sqrt{a})^2 = a\) for \(a \ge 0\).
However, for sums and differences, similar properties do not exist: the root of a sum is not equal to the sum of the roots. This is the most common mistake in this topic; the explanation is below.
| Property and condition | Example |
|---|---|
| √(a · b) = √a · √bfor a ≥ 0 and b ≥ 0 | √(4 · 25) = 2 · 5 = 10 |
| √(a / b) = √a / √bfor a ≥ 0 and b > 0 | √(81 / 16) = 9 / 4 |
| √(a²) = |a|for any a | √((−12)²) = 12 |
| (√a)² = afor a ≥ 0 | (√13)² = 13 |
| √(a² · b) = |a| · √bfor b ≥ 0 | √(25 · 3) = 5√3 |
How to extract a square root of a number
A calculator is not always at hand, and school assignments are almost always designed so that the root can be extracted. The procedure is as follows.
- Check the table of squares. If the number is in the table, the answer is ready: \(\sqrt{289} = 17\).
- Separate a round factor. For numbers with zeros, it's convenient to factor out \(100\), \(10,000\), etc.: \(\sqrt{6400} = \sqrt{64 \cdot 100} = 8 \cdot 10 = 80\).
- Factor into prime factors. This is a universal method: \(576 = 2^6 \cdot 3^2\), divide the exponents by two to get \(\sqrt{576} = 2^3 \cdot 3 = 24\).
- Convert a decimal fraction to a common fraction. \(\sqrt{1{,}44} = \sqrt{\dfrac{144}{100}} = \dfrac{12}{10} = 1{,}2\). This also helps avoid losing a decimal place.
- If there is no perfect square – the root is irrational. In this case, it's either left as a symbol in the answer (written as \(\sqrt{30}\)), or estimated between adjacent squares.
Checking takes a second: square the obtained number and see if you get the original radicand. \(24^2 = 576\) – correct.
Factoring out a multiplier from under the radical sign
A root is rarely left 'raw' – everything that can be extracted is factored out. The property \(\sqrt{a^2 b} = |a| \sqrt{b}\) for \(b \ge 0\) is used: look for the largest square factor of the radicand.
- \(\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}\);
- \(\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}\);
- \(\sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3}\).
The reverse operation is factoring a multiplier under the root: \(a\sqrt{b} = \sqrt{a^2 b}\) for \(a \ge 0\). For example, \(3\sqrt{5} = \sqrt{9 \cdot 5} = \sqrt{45}\).
Caution is needed with negative multipliers: the minus sign does not go under the root and remains outside. The correct notation is \(-2\sqrt{3} = -\sqrt{12}\), not \(\sqrt{12}\).
Why is this useful: after factoring out a multiplier, roots with the same radicand become like terms and can be added. Factoring a multiplier under the root is the quickest way to compare two expressions with roots.
What is the square root that cannot be extracted
If the radicand is not a perfect square, the root is an irrational number: its decimal representation is infinite and non-repeating. For example, \(\sqrt{2} \approx 1{,}4142\ldots\), and the exact value is written only with the symbol \(\sqrt{2}\). A rounded answer is given only when requested in the problem.
This root can be estimated without a calculator: find two adjacent perfect squares between which the radicand lies. The comparison rule applies: for \(a \ge 0\) and \(b \ge 0\), if \(a < b\), then \(\sqrt{a} < \sqrt{b}\) – a larger radicand corresponds to a larger root.
Examples of calculations with square roots
Example 1. Value of an expression with roots
Calculate \(\sqrt{121} - \sqrt{0{,}36} + \sqrt{\dfrac{9}{25}}\).
Step 1. \(\sqrt{121} = 11\), because \(11^2 = 121\).
Step 2. \(\sqrt{0{,}36} = 0{,}6\), because \(0{,}6^2 = 0{,}36\).
Step 3. Extract the root of the fraction for the numerator and denominator: \(\sqrt{\dfrac{9}{25}} = \dfrac{3}{5} = 0{,}6\).
Step 4. Add: \(11 - 0{,}6 + 0{,}6 = 11\).
Answer: 11.
Example 2. Root of a product
Calculate \(\sqrt{8} \cdot \sqrt{18}\).
Step 1. These roots cannot be extracted individually, so we combine them into one: \(\sqrt{8} \cdot \sqrt{18} = \sqrt{8 \cdot 18}\).
Step 2. Calculate the product under the root: \(8 \cdot 18 = 144\).
Step 3. \(\sqrt{144} = 12\).
Another approach is to factor out multipliers beforehand: \(\sqrt{8} = 2\sqrt{2}\) and \(\sqrt{18} = 3\sqrt{2}\), then \(2\sqrt{2} \cdot 3\sqrt{2} = 6 \cdot (\sqrt{2})^2 = 6 \cdot 2 = 12\). The results match.
Answer: 12.
Example 3. Extracting a root by factoring
Calculate \(\sqrt{576}\) and \(\sqrt{6400}\).
Step 1. The number 576 is not in the table of squares, so we factor it into prime factors: \(576 = 2^6 \cdot 3^2\).
Step 2. Divide the exponents by two: \(\sqrt{2^6 \cdot 3^2} = 2^3 \cdot 3 = 8 \cdot 3 = 24\).
Step 3. Check. \(24^2 = 576\) – correct.
Step 4. For the number 6400, it's more convenient to factor out a round number: \(\sqrt{6400} = \sqrt{64 \cdot 100} = 8 \cdot 10 = 80\). Check: \(80^2 = 6400\).
Answer: \(\sqrt{576} = 24\), \(\sqrt{6400} = 80\).
Example 4. Factoring out a multiplier and combining like terms
Simplify \(2\sqrt{98} - \sqrt{300} + \sqrt{72}\).
Step 1. Factor out multipliers from under each root: \(\sqrt{98} = \sqrt{49 \cdot 2} = 7\sqrt{2}\), so \(2\sqrt{98} = 14\sqrt{2}\).
Step 2. \(\sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3}\).
Step 3. \(\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}\).
Step 4. Only roots with the same radicand can be added: \(14\sqrt{2} + 6\sqrt{2} = 20\sqrt{2}\), and the term with \(\sqrt{3}\) remains separate.
Answer: \(20\sqrt{2} - 10\sqrt{3}\).
Example 5. Root of a square with a variable
Simplify \(\sqrt{(x-4)^2}\) for \(x < 4\).
Step 1. By the property of the root of a square, \(\sqrt{(x-4)^2} = |x-4|\) – immediately write the absolute value, not \(x - 4\).
Step 2. Remove the absolute value according to the condition. If \(x < 4\), then the difference \(x - 4\) is negative, and the absolute value of a negative number is its opposite: \(|x-4| = 4 - x\).
Step 3. Check. Let's take \(x = 1\): on the left, \(\sqrt{(1-4)^2} = \sqrt{9} = 3\), on the right, \(4 - 1 = 3\) – it matches.
Answer: \(4 - x\). For comparison, if \(x \ge 4\), the answer would be \(x - 4\).
Example 6. Comparing expressions with roots
Compare \(3\sqrt{5}\) and \(5\sqrt{2}\).
Step 1. Both numbers are positive, so we factor the multipliers under the root: \(3\sqrt{5} = \sqrt{9 \cdot 5} = \sqrt{45}\).
Step 2. \(5\sqrt{2} = \sqrt{25 \cdot 2} = \sqrt{50}\).
Step 3. Compare the radicands: \(45 < 50\), and a larger radicand corresponds to a larger root.
Answer: \(3\sqrt{5} < 5\sqrt{2}\).
Common mistakes when working with square roots
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Writing $\sqrt{a^2} = a$, forgetting the absolute value.
The correct equality is \(\sqrt{a^2} = |a|\). For a negative number, this is crucial: \(\sqrt{(-13)^2} = \sqrt{169} = 13\), not \(-13\). An arithmetic root cannot be negative, so the sign under the root 'straightens out'.
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Assuming that the root of a sum equals the sum of the roots: $\sqrt{9 + 16} = 3 + 4$.
There is no such property. First, perform the operation under the root: \(\sqrt{9 + 16} = \sqrt{25} = 5\), not 7. Properties exist only for products and quotients: \(\sqrt{ab} = \sqrt{a}\cdot\sqrt{b}\) and \(\sqrt{a/b} = \sqrt{a}/\sqrt{b}\).
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Answering that $\sqrt{49} = \pm 7$.
\(\sqrt{49} = 7\) – the arithmetic root is always unique and non-negative. Two values appear in a different problem: the equation \(x^2 = 49\) has roots \(x = 7\) and \(x = -7\). Do not confuse the notation for a root with solving an equation.
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Writing $\sqrt{-16} = -4$, reasoning 'minus times minus'.
A negative number cannot be under the root: \((-4)^2 = 16\), not \(-16\), so \(\sqrt{-16}\) has no meaning in real numbers. However, the expression \(-\sqrt{16} = -4\) is perfectly correct – the minus sign is before the root, not under it.
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Factoring a negative multiplier under the root: $-2\sqrt{3} = \sqrt{12}$.
The rule \(a\sqrt{b} = \sqrt{a^2 b}\) works only for \(a \ge 0\). The minus sign remains outside: \(-2\sqrt{3} = -\sqrt{12}\). Check the sign: \(-2\sqrt{3} \approx -3.46\), and \(\sqrt{12} \approx 3.46\) – the numbers are opposites.
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Losing a decimal place with a decimal fraction: $\sqrt{0{,}04} = 0{,}02$.
Check by squaring in reverse: \(0.02^2 = 0.0004\), it doesn't match. The correct answer is \(\sqrt{0.04} = 0.2\), since \(0.2^2 = 0.04\). A reliable method is to convert to a common fraction: \(\sqrt{\dfrac{4}{100}} = \dfrac{2}{10} = 0.2\).
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Being sure that the root is always less than the number itself.
This is true only for numbers greater than one. In the interval from 0 to 1, it's the opposite: \(\sqrt{0.25} = 0.5\), and \(0.5\) is greater than \(0.25\). For 0 and 1, the root equals the number itself.
Questions and answers
What is an arithmetic square root in simple terms?
It is a non-negative number that, when squared, gives the radicand. It is written as \(\sqrt{a}\), where the radicand must be non-negative: \(a \ge 0\). For example, \(\sqrt{64} = 8\), because \(8 \ge 0\) and \(8^2 = 64\).
How does a square root differ from an arithmetic square root?
A positive number has two square roots – opposite in sign. The arithmetic square root refers only to the non-negative one, and this is what the \(\sqrt{\phantom{a}}\) symbol denotes. Therefore, \(\sqrt{25} = 5\), although the equation \(x^2 = 25\) has two roots: 5 and \(-5\).
How to extract a square root of a number without a calculator?
First, check the number in the table of squares. If it's not there, factor it and extract perfect squares: \(784 = 16 \cdot 49\), so \(\sqrt{784} = 4 \cdot 7 = 28\). Always check the answer by the reverse operation: \(28^2 = 784\).
Is it possible to extract a square root of a negative number?
Not in real numbers: the square of any number is non-negative, so the expression \(\sqrt{-9}\) has no meaning. This is precisely the condition used to find the domain: for example, \(\sqrt{x-7}\) exists only when \(x \ge 7\).
What is the square root of a number squared?
The absolute value of that number: \(\sqrt{a^2} = |a|\). If \(a\) is non-negative, it's the number itself; if \(a\) is negative, it's its opposite: \(\sqrt{(-6)^2} = 6\).
How to factor a multiplier out from under the radical sign?
Find the largest square factor of the radicand and apply the property \(\sqrt{a^2 b} = |a|\sqrt{b}\). For example, \(\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}\). The reverse operation is factoring a multiplier under the root: \(3\sqrt{5} = \sqrt{45}\).
What is the square root of 2?
The number \(\sqrt{2}\) is irrational: its decimal representation is infinite and non-repeating, \(\sqrt{2} \approx 1.4142\). There is no exact value as a fraction, so it's left as the symbol \(\sqrt{2}\) in answers, and rounded only if the problem requires it.
How to compare two numbers with roots without calculating them?
Factor the multipliers under the root and compare the radicands: a larger radicand corresponds to a larger root. For example, compare \(\sqrt{17}\) and 4: since \(4 = \sqrt{16}\) and \(17 > 16\), then \(\sqrt{17} > 4\).
What grade are square roots studied in?
In 8th grade algebra classes: first, the definition of the arithmetic square root and its properties, then transformations of expressions with roots, and solving equations of the form \(x^2 = a\). Later, roots appear in the discriminant formula and in tasks for the OGE and USE exams.
