# Algebraic Surds

Contents

## LEARNING OBJECTIVES:

• To define algebraic surds
• To perform operations involving surds
• To simplify and rationalize surds

## MATH CONCEPTS

• SURDS- It is a root of a positive real quantity where its value cannot be exactly determined. It is an irrational number which cannot be simplified and usually in the form of $\sqrt[n]{x}$, n is the order of surd and x is called a radicand.. Examples of surds are $\sqrt{3}, \,\sqrt{11},\,\sqrt[3]{17}$
• IRRATIONAL NUMBER-  It is a real number that cannot be written as a simple fraction. All surds are irrational numbers but all irrational numbers are not surds.
• ADDITION AND SUBTRACTION OF SURDS

$a\sqrt{x}+c\sqrt{x}=(a+c)\sqrt{x}$,     $a\sqrt{x}-c\sqrt{x}=(a-c)\sqrt{x}$

• MULTIPLICATION AND DIVISION OF SURDS

$\sqrt{xy}=\sqrt{x}\times \sqrt{y}$,     $\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$

## IMPORTANCE

• Learning surds aids us to understand that we leave surds in root form as the values can’t be simplified. So when we do problem solving with surds, we usually try to rewrite the surds to more simplified forms and whenever necessary, we can take the approximate value of any surd up to any decimal to calculate.

## DISCUSSION

SURDS

As mentioned, surds are numbers with irrational values. Surds are examples of irrational numbers, whose ratio cannot be determined. Surds are usually enclosed in a radical sign with radicand. But remember, not all numbers enclosed in radical signs are surds because some of these can be simplified like square numbers.

More generally, we get a surd when we take the square root of a number that isn’t a square – so $\sqrt{3}$, $\sqrt{5}$ and $\sqrt{7}$ are all surds. Like many things in maths, surds can be simplified.

OPERATIONS INVOLVING SURDS

It is very important to understand how to add, subtract, multiply, and divide surds. In adding surds, you should remember this $a\sqrt{x}+c\sqrt{x}=(a+c)\sqrt{x}$  and  $a\sqrt{x}-c\sqrt{x}=(a-c)\sqrt{x}$. You can only add and subtract similar surds —- surds with the same index and radicand ($\sqrt{2}$ and $3\sqrt{2}$ are similar surds). If they are different, then all we can do is to leave it as it is($\sqrt{2}$ and $3\sqrt{5}$ are dissimilar surds because the radicands are different).

When multiplying and dividing surds, it is possible even the radicands are different but make sure that the indices are the same. Remember that when we multiply and divide surds, $\sqrt{xy}=\sqrt{x}\times \sqrt{y}$  and  $\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$

## ILLUSTRATIVE EXAMPLES

A) Find the sum or difference of the following surds.

1. $2\sqrt{5}+\sqrt{5}$
2. $4\sqrt{2}+3\sqrt{2}$
3. $2a\sqrt{b}+6a\sqrt{b}$
4. $10\sqrt{17}-8\sqrt{17}$
5. $21\sqrt{23}-6\sqrt{23}$
6. $50x\sqrt{39}-24x\sqrt{39}$

SOLUTIONS:

1. $2\sqrt{5}+\sqrt{5}=(2+1)\sqrt{5}=3\sqrt{5}$
2. $4\sqrt{2}+3\sqrt{2}=(4+3)\sqrt{2}=7\sqrt{2}$
3. $2a\sqrt{b}+6a\sqrt{b}=(2a+6a)\sqrt{b}=8a\sqrt{b}$
4. $10\sqrt{17}-8\sqrt{17}=(10-8)\sqrt{17}=2\sqrt{17}$
5. $21\sqrt{23}-6\sqrt{23}=(21-6)\sqrt{23}=15\sqrt{23}$
6. $50x\sqrt{39}-24x\sqrt{39}=(50x-24x)\sqrt{39}=26x\sqrt{39}$

B) Find the sum or difference of the following surds.

1. $\sqrt{2}\times \sqrt{3}$
2. $4\sqrt{5}\times 3\sqrt{2}$
3. $a\sqrt{2}\times 3a\sqrt{7}$
4. $\frac{\sqrt{10}}{\sqrt{2}}$
5. $\frac{\sqrt{26}}{\sqrt{13}}$
6. $\frac{6x\sqrt{2m}}{2x\sqrt{m}}$

SOLUTIONS:

1. $\sqrt{2}\times \sqrt{3}=\sqrt{(2)(3)}= \sqrt{6}$
2. $4\sqrt{5}\times 3\sqrt{2}=(4\times2)\sqrt{(5)(2)}= 8\sqrt{10}$
3. $a\sqrt{2}\times 3a\sqrt{7}=(a\times3a)\sqrt{(7)(2)}= 3a^{2}\sqrt{14}$
4. $\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{\frac{10}{2}}=\sqrt{5}$
5. $\frac{\sqrt{26}}{\sqrt{13}}=\sqrt{\frac{26}{13}}=\sqrt{2}$
6. $\frac{6x\sqrt{2m}}{2x\sqrt{m}}=\frac{6x}{2x}\sqrt{\frac{m}{n}}=3\sqrt{m}$

## SIMPLIFYING AND RATIONALISING SURDS

Simplifying surds means that we already factored out numbers and extracted the roots of it. Finding the factors and recognizing square numbers are the key concepts is simplifying surds. How will we know that a surd is already in its simplest form? The answer is this — a surd is already simplified if the radicand is a prime number or there is no square integer that can be a factor of the radicand. For example, $\sqrt{17}$ is already in simplest form because 17 is a prime number. Another one is $\sqrt{26}$, 26 has factors 2 and 13. Between 2 and 13, neither is a square number. Thus, $\sqrt{26}$ is already in simplest form.

Surds are considered irrational numbers, and having irrational numbers on the bottom of a fraction is not allowed. So, by completely applying the multiplication rule mentioned above, we can make the denominator rational. This process is called rationalisation of rationalising surds. For example, $\frac{2}{\sqrt{2}}$ needs to be rationalised because of the presence of $\sqrt{2}$ in the denominator. Do we need to rationalise $\frac{2}{\sqrt{2}}$? The answer is no because 2 is a rational number.

ILLUSTRATIVE EXAMPLES

A) Perform the given operation then simplify the following surds if necessary.

1. $\sqrt{18}+\sqrt{50}$
2. $\sqrt{32}+\sqrt{8}$
3. $\sqrt{48}-\sqrt{12}$
4. $\sqrt{28}-\sqrt{7}$
5. $\sqrt{45}-\sqrt{20}$
6. $\sqrt{96}+\sqrt{24}$

SOLUTIONS:

1. $\sqrt{18}+\sqrt{50}=\sqrt{9*2}+\sqrt{25*2}=3\sqrt{2}+5\sqrt{2}=8\sqrt{2}$
2. $\sqrt{32}+\sqrt{8}=\sqrt{16*2}+\sqrt{4*2}=4\sqrt{2}+2\sqrt{2}=6\sqrt{2}$
3. $\sqrt{48}-\sqrt{12}=\sqrt{16*3}-\sqrt{4*3}=4\sqrt{3}+2\sqrt{3}=6\sqrt{3}$
4. $\sqrt{28}-\sqrt{7}=\sqrt{4*7}-\sqrt{7}=2\sqrt{7}+\sqrt{7}=3\sqrt{7}$
5. $\sqrt{45}-\sqrt{20}=\sqrt{9*5}-\sqrt{4*5}=3\sqrt{5}+2\sqrt{5}=7\sqrt{5}$
6. $\sqrt{96}+\sqrt{24}=\sqrt{16*6}+\sqrt{4*6}=4\sqrt{6}+2\sqrt{6}=6\sqrt{6}$

1. $\sqrt{3}*\sqrt{6}$
2. $5\sqrt{2}*4\sqrt{6}$
3. $(2+\sqrt{10})*(2-\sqrt{10})$
4. $\frac{\sqrt{20}}{\sqrt{5}}$
5. $\frac{6\sqrt{12}}{2\sqrt{3}}$
6. $\frac{\sqrt{300}}{\sqrt{2}}$

SOLUTIONS:

1. $\sqrt{3}*\sqrt{6}=\sqrt{18}=\sqrt{(9)(2)}=3\sqrt{2}$
2. $5\sqrt{2}*4\sqrt{6}=(5)(2)\sqrt{(2)(6)}=10\sqrt{12}=10\sqrt{(4)(3)}=(10)(2)\sqrt{3}=20\sqrt{3}$
3. $(2+\sqrt{10})*(2-\sqrt{10})=(2)(2)-2\sqrt{10}(10)(-10)=4-100=-96
$
4. $\frac{\sqrt{20}}{\sqrt{5}}=\sqrt{\frac{20}{5}}=\sqrt{4}=2$
5. $\frac{6\sqrt{12}}{2\sqrt{3}}=\frac{6}{3}\sqrt{\frac{12}{3}}=2\sqrt{4}=4$
6. $\frac{\sqrt{300}}{\sqrt{2}}=\sqrt{\frac{300}{2}}=\sqrt{150}=\sqrt{(25)(6)}=5\sqrt{6}$

C) Rationalise the following surds.

1. $\frac{4}{9\sqrt{2}}$
2. $\frac{5}{1+\sqrt{2}}$

SOLUTIONS:

1. $\frac{4}{9\sqrt{2}}= \frac{4}{9\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{4\sqrt{2}}{9\sqrt{2}(\sqrt{2})}=\frac{4\sqrt{2}}{9\sqrt{4}}=\frac{4\sqrt{2}}{9(2)}=\frac{4\sqrt{2}}{18}=\frac{2\sqrt{2}}{9}$
2. $\frac{5}{1+\sqrt{2}}$

$=\frac{5}{1+\sqrt{2}}*\frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{5(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}=\frac{5-5\sqrt{2}}{1-(\sqrt{4})}=\frac{5-5\sqrt{2}}{1-(\pm 2)}$

$\frac{5-5\sqrt{2}}{1-(\pm 2)}=\frac{5-5\sqrt{2}}{1-2}=-5+5\sqrt{2}$  or

$\frac{5-5\sqrt{2}}{1-(\pm 2)}=\frac{5-5\sqrt{2}}{1+2}=\frac{5-5\sqrt{2}}{3}$