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1211 LearnersLast updated on December 2, 2025

Surd is a term that we use to refer to square roots of non-perfect squares. Surds also include higher roots, such as cube roots, that cannot be simplified into rational numbers. In this topic, we are going to learn more about surds and their various types.

A surd is a mathematical term used to describe irrational numbers that can be expressed as the root of an integer. When a root cannot be simplified further, we call that a surd. For example, √4 is not a surd because it can be simplified to 2. When we simplify √4, we get two because the square root of 4 is 2. Surds help in keeping calculations exact rather than using decimal approximations.
Here are some surds examples:
\(โ2, โ3, โ5, and โ7\)are surds because they cannot be simplified further.
√16 is not a surd because it simplifies to 4.
Using surds in calculations yields precise results rather than approximate decimals, which is helpful in algebra and higher math.
Surds are square roots that cannot be simplified into regular numbers. Knowing their rules makes math easier and more fun!
Example:
\(โ(9 + โ16) = 3 + 2, so โ(9 โ โ16) = 3 โ 2.\)
Example:
\(โโโโโโโ2โ3 + 5โ3 = 7โ3,\) but √2 + √3 cannot be combined.
Example:
√2 × √8 = √(2 × 8) = √16 = 4.
Example:
\(โ8 รท โ2 = โ(8 รท 2) = โ4 = 2.\)
We can classify surds into six different types:


There are 6 rules that we use for the calculation of surds:
Rule 1: \(\sqrt{a\times b} = \sqrt a \times \sqrt b\)
Example: \(\sqrt{36} = \sqrt{9ร4} = \sqrt9 ร \sqrt4 = 3 ร 2 = 6\)
Rule 2: \({\sqrt a \over \sqrt b} = \sqrt{a \over b}\)
Example: \({\sqrt{18} \over \sqrt2} = {\sqrt{18\over2}} = \sqrt9 = 3\)
Rule 3: \({b\over \sqrt a} = {{b\over \sqrt a} \times {\sqrt a\over \sqrt a}} = {b \sqrt a \over \ a}\)
Example: \({3\over \sqrt 5} = {{3\over \sqrt 5} \times {\sqrt 5\over \sqrt 5}} = {3 \sqrt 5 \over \ 5}\)
Rule 4: \(a \sqrt c \space \pm \space b\sqrt c = (a \space \pm \space b)\times \sqrt c\)
Example: \(5 \sqrt 5 \space \pm \space 3\sqrt 5 = (5 \space \pm \space 3)\times \sqrt 5\)
Rule 5: \({c \over {a\space+\space b\sqrt n}} = {{c \over {a\space+\space b\sqrt n}}} \times {{a\space-\space b\sqrt n}\over {a\space-\space b\sqrt n}} = {c \times ({a\space-\space b\sqrt n)}\over {a^2\space-\space b^2 n}}\)
Example: \({5\over {4\space+\space 2\sqrt 3}} = {{5 \over {4\space+\space 2\sqrt 3}}} \times {{4\space-\space 2\sqrt 3}\over {4\space-\space 2\sqrt 3}} = {5 \times ({4\space-\space 3\sqrt 2)}\over {4^2\space-\space (2^2 \times 3)}} = {20-15 \sqrt 2\over 4}\)
Rule 6: \({c \over {a\space-\space b\sqrt n}} = {{c \over {a\space-\space b\sqrt n}}} \times {{a\space+\space b\sqrt n}\over {a\space+\space b\sqrt n}} = {c \times ({a\space+\space b\sqrt n)}\over {a^2\space-\space b^2 n}}\)
Example: \({5\over {4\space-\space 2\sqrt 3}} = {{5 \over {4\space-\space 2\sqrt 3}}} \times {{4\space+\space 2\sqrt 3}\over {4\space+\space 2\sqrt 3}} = {5 \times ({4\space+\space 3\sqrt 2)}\over {4^2\space-\space (2^2 \times 3)}} ={20+15 \sqrt 2\over 4} \)
When solving for surds, there are a few steps we need to look out during each operation:
These are some of the few ways we can solve for surds when using the basic arithmetic operations.
Surds WorkSheet:
Identify Surds
√5
√16
√2
√49
√11
Working with surds can feel tricky at first, but with a few simple tips, it becomes much easier. These tricks are also helpful for parents and teachers to guide children and simplify surds in a fun, easy way.
Example:
\(โ180 = โ(2 ร 2 ร 3 ร 3 ร 5) = (2 ร 3)โ5 = 6โ5\)
This trick makes simplification of surds easier for kids and for indies learning on their own.
Students tend to make mistakes while learning surds. Being aware of such mistakes can work in our favor. Take a look at some of the most common mistakes and ways to avoid them:
Surds are used in fields where precise calculations involving irrational numbers are required:
Simplify โ72.
6√2
Factorize 72, and you will get \({\sqrt{36 \times 2}} = \sqrt{36} \times \sqrt 2\)
Since \(\sqrt{36}\) = 6 we get 6√2.
Add 3โ5 +7โ5
10√5
Since both terms have the same surd √5, we will add the coefficients: \(3 + 7 = 10\)
So, \(3โ5 + 7โ5 = 10โ5\)
8โ3 - 2โ3
6√3
So, \(8โ3 - 2โ3 = 6โ3\)
Divide: โ48/โ3
4
\({\sqrt {48} \over \sqrt {3} } = {\sqrt{48\over 3}} = \sqrt {16} = 4\)
Solve: (โ18 + โ8) / โ2
5
\(\sqrt {18} = 3\sqrt 2, \sqrt8 = 2\sqrt 2\)
Sum of \(\sqrt {18} \space\text{and} \sqrt8 \) : \(x3โ2 + 2โ2 = 5โ2\)
Dividing by\(ย โ2:ย 5โ2ย รทย โ2 = 5\)
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.






