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Surds
In this article, we will learn about
- How to identify surds
- Simplify expressions that involve surds
This article is specifically written to meet the requirements of Secondary 3 Additional Mathematics.
What is a Surd?
Surds are essentially irrational numbers that comprise the square root or cube root of a number.
| Surds | Not Surds |
|---|---|
| \(\sqrt{5}\) | \(\sqrt{5}=25\) |
| \(\sqrt[3] {2}\) | \(\sqrt[3] {8} =2\) |
| \(\begin{align*} \frac {1+\sqrt{2}}{5} \end{align*}\) | \(\sqrt{4} \;+\; \sqrt[3]{27} = 5\) |
In this article, we will focus only on surds that involve square roots like \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\), \(\sqrt{10}\), \(\sqrt{11}\).
Question 1:
Identify the surd in the following
A. \(\begin{align*} \frac {\sqrt{100}+\sqrt{9}}{2} \end{align*}\)
B. \(\begin{align*} \frac {\sqrt{7}+\sqrt{4}}{5} \end{align*}\)
C. \(\begin{align*} \sqrt[3]{27}+1 \end{align*}\)
D. \(\begin{align*} 2\sqrt {4} \end{align*}\)
Solution:
B) \(\begin{align*} \frac {\sqrt{7}+\sqrt{4}}{5} \end{align*}\)
Explanation:
The option (B) is a surd because upon solving the equation, it gives an irrational value and a long string of Decimals.
Essentially, when you calculate the value of \(\sqrt{2}\) it is equal to \(1.4141414…..\)
In its Square root form, the quantity is called surd and the decimal form is called an irrational number.
Laws Governing Surds
Let us perform a few sets of exercises to understand how the simple operations around surds work.
| TRUE / FALSE | |||
|---|---|---|---|
| Addition | \(\begin{align*} \sqrt {5+3} = 2.828 \end{align*}\) | \(\begin{align*} \sqrt {5}+\sqrt {3} = 3.968 \end{align*} \) | FALSE |
| Subtraction | \(\begin{align*} \sqrt {5-3} = 1.414 \end{align*}\) | \(\begin{align*} \sqrt {5}-\sqrt {3} = 0.504 \end{align*} \) | FALSE |
| Multiplication | \(\begin{align*} \sqrt {5\times3} =3.872 \end{align*}\) | \(\begin{align*} \sqrt {5}\times\sqrt {3} = 3.872 \end{align*} \) | TRUE |
| Division | \(\begin{align*} \sqrt {\frac{5}{3}} = 1.290 \end{align*}\) | \(\begin{align*} \frac{\sqrt {5}}{\sqrt {3}} = 1.290 \end{align*} \) | TRUE |
| Square | \(\begin{align*} \sqrt {5^2} = 5 \end{align*}\) | \(\begin{align*} (\sqrt {5})^2 = 5 \end{align*}\) | TRUE |
From this table, we observe that for addition and subtraction, it does not work but it works for multiplication, division, and square.
From the above exercise, we can now look at the law of surds.
If \(a > 0, b > 0\) then
| Multiplication | \(\begin{align*} \sqrt {a\times b} &= \sqrt {a} \,\times\, \sqrt {b} \end{align*}\) |
|---|---|
| Division | \(\displaystyle{\sqrt {\frac {a}{b}} = \frac {\sqrt a}{\sqrt b}}\) |
| Square | \(\sqrt {a} \,\times \sqrt {a} = (\sqrt a)^2 =a\) |
Example 1:
Simplify each of the following
- \(\sqrt18\)
- \(\displaystyle{\sqrt 18 \over \sqrt 2}\)
- \(\sqrt2 \times \sqrt18\)
Solution:
1. \(\sqrt{18}\) can be written as \(\sqrt {6} \times \sqrt {3} \;\; \text{&}\;\; \sqrt {9} \times \sqrt {2}\)
However, in the first case \(\sqrt {6} \times \sqrt {3}\,,\) when we simplify they can only give an irrational form and not a simplified surd form.
The second case, however, \(\sqrt {9} \times \sqrt {2}\) can be simplified to, \(3\sqrt {2}\) and hence, it is the simplest form available.
2. \(\begin{align} \frac {\sqrt {18}}{\sqrt {2}} = \sqrt {\frac {18}{2}} = \sqrt {9} =3 \end{align}\)
In this, we can unify the roots into one under the Law governing Surds. The further cancellation results in a perfect square which when under root gives the result 3.
3. \(\sqrt {2} \times \sqrt {18} = \sqrt {2\times 18} = \sqrt {36} =6\)
In this, we can unify the roots into one under the Law governing Surds. The further cancellation results in a perfect square which when under root gives the result 6.
Question 2:
Simplify \(\sqrt{12}\) without using a calculator.
Solution:
\(\sqrt {12} = \sqrt {3\times4} = \sqrt {3} \;\times\; \sqrt {4} = \sqrt {3} \;\times 2 = 2\sqrt{3}\)
Example 2:
Simplify \(\sqrt {50} + \sqrt {8}\) without using a calculator.
Solution:
\(\begin{align*} &\sqrt {50} + \sqrt {8}\\[2ex] &= \sqrt {2\times 25} + \sqrt {2\times 4} \\[2ex] &= \sqrt {2} \times \sqrt {25} + \sqrt {2} \times \sqrt {4}\\ &\quad \small{\text {( split up using the multiplication rule in Law of Surds)}} \\[2ex] &= \sqrt {2} \times 5 + \sqrt {2} \times 2 \\[2ex] &= 5\sqrt {2} + 2\sqrt {2} \\[2ex] &= 7\sqrt {2} \end{align*}\)
Question 3:
Simplify \(\sqrt {20} + \sqrt {45}\) without using a calculator.
Solution:
\(\begin{align*} &\sqrt {20} + \sqrt {45}\\[2ex] &= \sqrt {4\times 5} + \sqrt {5\times 9}\\[2ex] &= \sqrt {4} \times \sqrt {5} + \sqrt {5} \times \sqrt {9}\\ &\quad \small{\text{( split up using the multiplication rule in Law of Surds)}} \\[2ex] &= 2\sqrt {5} + 3\sqrt {5} \\[2ex] &= 5\sqrt {5} \end{align*} \)
Example 3:
Simplify \((\,5-3\sqrt{2}\, )(\,1-\sqrt{2}\,)\) without using a calculator.
Solution:
Using the Distributive method or the “Rainbow method”, we expand the brackets
\(\begin{align*} & (\,5-3\sqrt{2}\, )(\,1-\sqrt{2}\,) \\[2ex] & = 5-5\sqrt {2} -3\sqrt{2} + (\,3\sqrt {2}\,)(\,\sqrt {2}\,)\\[2ex] & = 5 - 8\sqrt{2} +6 \\[2ex] & = 11 - 8\sqrt{2} \end{align*}\)
Question 3:
Simplify \((\,1+\sqrt{5} \;)(\,2-3\sqrt5\,)\) without using a calculator.
Solution:
Using the Distributive method or the “Rainbow method”, we expand the brackets
\(\begin{align*} &(\,1+\sqrt{5} \;)(\,2-3\sqrt5\,) \\ \\ &= 2-3\sqrt{5} +2\sqrt{5}-3 \times 5 \\ \\ &= 2-\sqrt{5}-15 \\ \\ &= -13-\sqrt{5} \end{align*}\)
Rationalising the Denominator of a Surd
A fraction with a surd in its denominator can be simplified by making the denominator a rational number.
You are supposed to rationalise a denominator by multiplying the numerator and denominator of a fraction with the same surd in the denominator.
Example 4:
Simplify the following by rationalising the denominator.
\(\displaystyle{\frac {5}{\sqrt{3}}}\)
Solution:
\(\begin{align*} \frac {5}{\sqrt{3}} &= \frac {5 \times \sqrt{3}} {\sqrt{3}\times \sqrt{3}} \\ \\ &= \frac {5\sqrt{3}}{3} \end{align*}\)
Conclusion
In this article, we have learnt how to identify Surds, which are defined as an irrational number that comprises the square root or the cube root. We have also learnt how to simplify expressions involving surds and how to split the composite number such that one of the numbers is a perfect square. All the topics we have discussed here are as per the Secondary 3 Additional Mathematics syllabus.
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