Dividing, Multiplying and Rationalizing Surds number

25 is perfect square as$\sqrt{25} =\hspace{0.17em}5$, 8 is a perfect cube since $\sqrt[3]{8 } =\hspace{0.17em}2$, Most numbers do not have exact roots. Such numbers are not perfect squares, cubes and etc and it becomes impossible to find their roots exactly such as roots are called surds or irrational numbers. In this article we are going to see how Dividing,Multiplying and Rationalizing is done with surds numbers.

Example of surds

$\sqrt{2 }\sqrt{3} \sqrt{5 } \sqrt[8]{5}$

Numbers with exact roots can be written in the form$\frac{a}{b}$where a and b are integers with a  and  HCF(a,b)=1

 

Multiplying  surds

Consider the

$\sqrt{9} and \sqrt{4}$ $\sqrt{9}X\sqrt{4} =\hspace{0.17em}\sqrt{9x4} = \sqrt{36} =6$

Generally  

$\sqrt{axb} = \sqrt{a}.\sqrt{b}$

 

 

Express the following as a simple possible surds

1. $\sqrt{4x3} = \sqrt{4}x\sqrt{3} = 2\sqrt{3}$
2. $\sqrt{48} = \sqrt{16x3} =4\sqrt{3}$
3. $\sqrt{75} = \sqrt{25x3} = 5\sqrt{3}$
4. $\sqrt{48} + \sqrt{75} = 4\sqrt{3} + 5\sqrt{3} = 9\sqrt{3}$

 

Expand

1.

$(a +\sqrt{b})(a–\sqrt{b})$

=

$a^2+–a\sqrt{b} +a\sqrt{b} –b^2$

=

$a^2 – b^2$

2.

$(a+\sqrt{b})(c+\sqrt{a})$

=

$ac+a\sqrt{a}+c\sqrt{b}+a\sqrt{a}$

 

Dividing surds

Consider perfect squares 81 and 25

$\sqrt{81} =9, \sqrt{25} =5$

=

$\sqrt{\frac{81}{25}} = \frac{9}{5}$

Generally 

$\sqrt{\frac{a}{b}} =\frac{\sqrt{a}}{\sqrt{b}}$

 

Conjugate surds

Let a, b

$\in IR$

with b,

$b>0$

.

Consider

$a+\sqrt{b} and a–\sqrt{b}$

 

$$\begin{gathered} (a+\sqrt{b})(a–\sqrt{b}) \\ =a^2–a\sqrt{b}+a\sqrt{b}–b^2 \\ =a^2–b^2 \end{gathered}$$

$a^2–b^2$ is a rational number thus, the product of the surd and it’s conjugate is a rational number

Example

$$\begin{gathered} 3+\sqrt{5} \\ =(3+\sqrt{5})(3–\sqrt{5}) \\ =9–3\sqrt{5}+3\sqrt{5}–5 \\ =9–5 \\ =4 \end{gathered}$$

Rationalizing Surds

To simplify expressions containing surd in the denomination, we rationalize the denominator. This is done by multiplying both the denominator and numerator by the conjugate surd of the denominator.

Examples

Rationalize the following surds

1. $\frac{1}{3–\sqrt{5}}$
2. $\frac{3–2\sqrt{5}}{4+3\sqrt{5}}$
3. $\frac{\sqrt{3}–\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Solutions

1.

 

$$\begin{gathered} \frac{1}{3–\sqrt{5}} \\ =\frac{3+\sqrt{5}}{(3–\sqrt{5})(3+\sqrt{5})} \\ =\frac{3+\sqrt{5}}{9+3\sqrt{5}–3\sqrt{5}–5} \\ =\frac{3+\sqrt{5}}{4} \end{gathered}$$

 

2. $$\begin{gathered} \frac{3–2\sqrt{5}}{4+3\sqrt{5}} \\ =\frac{(3–2\sqrt{5})(4–3\sqrt{5})}{(4+3\sqrt{5})(4–3\sqrt{5})} \\ =\frac{12–9\sqrt{5}–8\sqrt{5}+6}{16–12\sqrt{5}+12\sqrt{5}–9\left(5\right)} \\ =\frac{12–17\sqrt{5}+30}{16–45} \\ =\frac{–42+17\sqrt{5}}{29} \end{gathered}$$

3 $$\begin{gathered} \frac{\sqrt{3}–\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{(\sqrt{3}–\sqrt{2})(\sqrt{3}–\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}–\sqrt{2})} \\ =\frac{3–\sqrt{6}–\sqrt{6}+2}{3–\sqrt{6}–\sqrt{6}–2} \\ =\frac{3–2\sqrt{6}+2}{3–2} \\ =5–2\sqrt{6} \end{gathered}$$

Equality of surds

Two surds  $a+\sqrt{b} and c +\sqrt{d}$ are equal if a=c and b=d.

Example

$14+\sqrt{96}$

Solution

 

$$\begin{gathered} \sqrt{14}+\sqrt{96 }\hspace{0.17em}= (\sqrt{x}+\sqrt{xy})(\sqrt{xy}+\sqrt{y}) \\ =x+\sqrt{xy}+\sqrt{xy}+y \\ =x+y+2\sqrt{xy} \\ 14+\sqrt{96}=x+y+2\sqrt{xy} \\ =x+y =\hspace{0.17em}14 ......\left(1\right) \\ =\sqrt{xy}=\sqrt{96} .....\left(2\right) \\ =\frac{4xy}{4}=\frac{96}{4} ......\left(3\right) \\ from 1 y= 14–x sub in \left(3\right) \\ x(14–x)=\frac{96}{4} \\ 4x–x^2 = \frac{96}{4} \\ 14x–x^2=24 \\ {x}^2–14x–24 =0, \\ (x–12)(x–2) =0 \\ when x=2, y=12 \\ when x=12, y=2 \\ therefore \sqrt{14}+\sqrt{96} =\sqrt{2}+\sqrt{12} \end{gathered}$$

 

Exercise

1. find the square root of $\sqrt{24–16\sqrt{2}}$
2. use a calculator to evaluate $\sqrt{5+2\sqrt{6}} and \sqrt{2}–\sqrt{3}$,what can you conclude
3. find the values of $\sqrt{24+2x}=4$

comment  your answers below