How to rationalise the denominator of .1a+b

To rationalise the denominator of 1a+b, we multiply by the conjugate of (a+b). The conjugate of (a+b) is (a−b). So, we multiply by (a−b) in the numerator and denominator both to make the denominator rational. 1(a−b)×(a+b)(a+b) = (a+b)a2−(b)2 = a+ba2−b

Exercise 1.5 ML Aggarwal Class 9 ICSE Mathematics Chapter 1 Rational and Irrational Numbers Best Solutions

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Chapter Test-1

Exercise 1.5 ML Aggarwal Class 9 Mathematics ICSE Textbook contains a total of 12 questions. The questions are based on the topic: Rationalisation of surds.

Exercise 1.5 ML Aggarwal Class 9 Mathematics Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12

The first question of exercise 1.5 ML Aggarwal Class 9 asks us to rationalise the denominator of the given surd expressions.

1. Rationalise the denominator of the following:

(i) $\frac{3}{4 \sqrt{5}}$

Answer

To rationalise the denominator, we multiply with a rationalising factor of the denominator in both numerator and denominator.

The rationalising factor of $4 \sqrt{5}$ is $\sqrt{5}$ .

Multiplying by $\sqrt{5}$ in numerator and denominator both, we get.

$∴ \frac{3}{4 \sqrt{5}} = \frac{3 \times \sqrt{5}}{4 \sqrt{5} \times \sqrt{5}} = \frac{3 \sqrt{5}}{4 \times 5} =$ $\frac{3 \sqrt{5}}{20}$ .

(ii) $\frac{5 \sqrt{7}}{\sqrt{3}}$

Answer

The rationalising factor of $\sqrt{3}$ is $\sqrt{3}$ .

Multiplying by $\sqrt{3}$ in numerator and denominator both, we get.

$∴ \frac{5 \sqrt{7}}{\sqrt{3}} = \frac{5 \sqrt{7} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{5 \sqrt{7 \times 3}}{\sqrt{3 \times 3}} =$ $\frac{5 \sqrt{21}}{3}$ .

(iii) $\frac{3}{4 - \sqrt{7}}$

Answer

The rationalising factor of ( $4 - \sqrt{7}$ ) is ( $4 + \sqrt{7}$ ).

Multiplying by $4 + \sqrt{7}$ in numerator and denominator both, we get.

$∴ \frac{3}{4 - \sqrt{7}}$

$= \frac{3 \times (4 + \sqrt{7})}{(4 - \sqrt{7}) (4 + \sqrt{7})}$

$= \frac{12 + 3 \sqrt{7}}{4^{2} - {(\sqrt{7})}^{2}}$

$= \frac{12 + 3 \sqrt{7}}{16 - 7}$

$= \frac{3 (4 + \sqrt{7})}{9}$

= $\frac{(4 + \sqrt{7})}{3}$ .

(iv) $\frac{17}{3 \sqrt{2} + 1}$

Answer

The rationalising factor of ( $3 \sqrt{2} + 1$ ) is ( $3 \sqrt{2} - 1$ ).

Multiplying by $3 \sqrt{2} - 1$ in numerator and denominator both, we get.

$∴ \frac{17}{3 \sqrt{2} + 1}$

$= \frac{17}{3 \sqrt{2} + 1} \times \frac{3 \sqrt{2} - 1}{3 \sqrt{2} - 1}$

$= \frac{17 (3 \sqrt{2} - 1)}{{(3 \sqrt{2})}^{2} - 1^{2}}$

$= \frac{17 (3 \sqrt{2} - 1)}{18 - 1}$

$= \frac{17 (3 \sqrt{2} - 1)}{17}$

= $(3 \sqrt{2} - 1)$ .

(v) $\frac{16}{\sqrt{41} - 5}$

Answer

The rationalising factor of $(\sqrt{41} - 5)$ is $(\sqrt{41} + 5)$ .

Multiplying by $(\sqrt{41} + 5)$ in numerator and denominator both, we get.

$∴ \frac{16}{(\sqrt{41} - 5)}$

$= \frac{16}{(\sqrt{41} - 5)} \times \frac{(\sqrt{41} + 5)}{(\sqrt{41} + 5)}$

$= \frac{16 \times (\sqrt{41} + 5)}{{(\sqrt{41})}^{2} - 5^{2}}$

$= \frac{16 (\sqrt{41} + 5)}{41 - 25}$

$= \frac{16 (\sqrt{41} + 5)}{16}$

= $(\sqrt{41} + 5)$ .

(vi) $\frac{1}{\sqrt{7} - \sqrt{6}}$

Answer

The rationalising factor of $(\sqrt{7} - \sqrt{6})$ is $(\sqrt{7} + \sqrt{6})$ .

Multiplying by $(\sqrt{7} + \sqrt{6})$ in numerator and denominator both, we get.

$∴ \frac{1}{(\sqrt{7} - \sqrt{6})}$

$= \frac{1}{(\sqrt{7} - \sqrt{6})} \times \frac{(\sqrt{7} + \sqrt{6})}{(\sqrt{7} + \sqrt{6})}$

$= \frac{(\sqrt{7} + \sqrt{6})}{{(\sqrt{7})}^{2} - {(\sqrt{6})}^{2}}$

$= \frac{(\sqrt{7} + \sqrt{6})}{7 - 6}$

= $(\sqrt{7} + \sqrt{6})$

(vii) $\frac{1}{\sqrt{5} + \sqrt{2}}$

Answer

The rationalising factor of $(\sqrt{5} + \sqrt{2})$ is $(\sqrt{5} - \sqrt{2})$ .

Multiplying by $(\sqrt{5} + \sqrt{2})$ in numerator and denominator both, we get.

$∴ \frac{1}{(\sqrt{5} + \sqrt{2})}$

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

$= \frac{(\sqrt{5} + \sqrt{2})}{{(\sqrt{5})}^{2} - {(\sqrt{2})}^{2}}$

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

= $\frac{(\sqrt{5} + \sqrt{2})}{3}$ .

(viii) $\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}$

Answer

The rationalising factor of $(\sqrt{2} - \sqrt{3})$ is $(\sqrt{2} + \sqrt{3})$ .

Multiplying by $(\sqrt{2} + \sqrt{3})$ in numerator and denominator both, we get.

$∴ \frac{(\sqrt{2} + \sqrt{3})}{(\sqrt{2} - \sqrt{3})}$

$= \frac{(\sqrt{2} + \sqrt{3})}{(\sqrt{2} - \sqrt{3})} \times \frac{(\sqrt{2} + \sqrt{3})}{(\sqrt{2} + \sqrt{3})}$

$= \frac{{(\sqrt{2} + \sqrt{3})}^{2}}{{(\sqrt{2})}^{2} - {(\sqrt{3})}^{2}}$

$= \frac{{(\sqrt{2})}^{2} + {(\sqrt{3})}^{2} + 2 \times \sqrt{2} \times \sqrt{3}}{2 - 3}$

$= \frac{2 + 3 + 2 \sqrt{6}}{- 1}$

= $- (5 + 2 \sqrt{6})$

= $- 5 - 2 \sqrt{6}$ .

The second question of exercise 1.5 ML Aggarwal Class 9 asks us to simplify the given surd expressions by rationalising the denominator:

2. Simplify each of the following by rationalising the denominator:

(i) $\frac{7 + 3 \sqrt{5}}{7 - 3 \sqrt{5}}$

Answer

The rationalising factor of $(7 - 3 \sqrt{5})$ is $(7 + 3 \sqrt{5})$ .

Multiplying by $(7 + 3 \sqrt{5})$ in numerator and denominator both, we get.

$\frac{7 + 3 \sqrt{5}}{7 - 3 \sqrt{5}}$ = $\frac{7 + 3 \sqrt{5}}{7 - 3 \sqrt{5}} \times \frac{7 + 3 \sqrt{5}}{7 + 3 \sqrt{5}}$

= $\frac{{(7 + 3 \sqrt{5})}^{2}}{7^{2} - {(3 \sqrt{5})}^{2}}$

= $\frac{7^{2} + {(3 \sqrt{5})}^{2} + 2 \times 7 \times (3 \sqrt{5})}{49 - 45}$

= $\frac{49 + 45 + 42 \sqrt{5}}{4}$

= $\frac{94 + 42 \sqrt{5}}{4}$

= $\frac{2 (47 + 21 \sqrt{5})}{4}$

= $\frac{47 + 21 \sqrt{5}}{2}$ .

(ii) $\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}}$

Answer

The rationalising factor of the denominator $3 + 2 \sqrt{2}$ is $3 - 2 \sqrt{2}$ .

Multiplying by $3 - 2 \sqrt{2}$ in numerator and denominator both, we get.

$\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}}$ = $\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}} \times \frac{3 - 2 \sqrt{2}}{3 - 2 \sqrt{2}}$

= $\frac{{(3 - 2 \sqrt{2})}^{2}}{3^{2} - {(2 \sqrt{2})}^{2}}$

= $\frac{3^{2} + {(2 \sqrt{2})}^{2} + 2 \times 3 \times (2 \sqrt{2})}{9 - 8}$

= $\frac{9 + 8 + 12 \sqrt{2}}{1}$

= $17 + 12 \sqrt{2}$ .

(iii) $\frac{5 - 3 \sqrt{14}}{7 + 2 \sqrt{14}}$

Answer

The rationalising factor of the denominator $7 + 2 \sqrt{14}$ is $7 - 2 \sqrt{14}$ .

Multiplying by $7 - 2 \sqrt{14}$ in numerator and denominator both, we get.

$\frac{5 - 3 \sqrt{14}}{7 + 2 \sqrt{14}}$ = $\frac{5 - 3 \sqrt{14}}{7 + 2 \sqrt{14}} \times \frac{7 - 2 \sqrt{14}}{7 - 2 \sqrt{14}}$

= $\frac{(5 - 3 \sqrt{14}) (7 - 2 \sqrt{14})}{7^{2} - {(2 \sqrt{14})}^{2}}$

= $\frac{5 (7 - 2 \sqrt{14}) - 3 \sqrt{14} (7 - 2 \sqrt{14})}{49 - 56}$

= $\frac{35 - 10 \sqrt{14} - 21 \sqrt{14} + 6 \sqrt{14 \times 14}}{- 7}$

= $\frac{35 - 31 \sqrt{14} + 84}{- 7}$

= $\frac{119 - 31 \sqrt{14}}{- 7}$

= $\frac{- 119 + 31 \sqrt{14}}{7}$ .

The third question of exercise 1.5 ML Aggarwal Class 9 asks us to simplify the given expression.

3. Simplify: $\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3 \sqrt{2}}{\sqrt{15} + 3 \sqrt{2}}$ .

Answer

We have

$\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3 \sqrt{2}}{\sqrt{15} + 3 \sqrt{2}}$

Multiplying by rationalising factors of the denominators of each terms, we get.

$\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} \times \frac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} - \frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} - \frac{3 \sqrt{2}}{\sqrt{15} + 3 \sqrt{2}} \times \frac{\sqrt{15} - 3 \sqrt{2}}{\sqrt{15} - 3 \sqrt{2}}$

$= \frac{7 \sqrt{30} - 21}{{(\sqrt{10})}^{2} - {(\sqrt{3})}^{2}} - \frac{2 \sqrt{30} - 10}{{(\sqrt{6})}^{2} - {(\sqrt{5})}^{2}} - \frac{3 \sqrt{30} - 18}{{(\sqrt{15})}^{2} - {(3 \sqrt{2})}^{2}}$

$= \frac{7 \sqrt{30} - 21}{10 - 3} - \frac{2 \sqrt{30} - 10}{6 - 5} - \frac{3 \sqrt{30} - 18}{15 - 18}$

$= \frac{7 (\sqrt{30} - 3)}{7} - \frac{(2 \sqrt{30} - 10)}{1} + \frac{3 (\sqrt{30} - 6)}{3}$

$= (\sqrt{30} - 3) - (2 \sqrt{30} - 10) + (\sqrt{30} - 6)$

$= \sqrt{30} - 3 - 2 \sqrt{30} + 10 + \sqrt{30} - 6$

= $1$

The fourth question of exercise 1.5 ML Aggarwal Class 9 asks us to simplify the given surd expressions.

4. Simplify: $\frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}$ .

Answer

We have

$\frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}$

Multiplying by rationalising factors of the denominators of each terms, we get.

$= \frac{1}{\sqrt{4} + \sqrt{5}} \times \frac{\sqrt{4} - \sqrt{5}}{\sqrt{4} - \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} \times \frac{\sqrt{5} - \sqrt{6}}{\sqrt{5} - \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} \times \frac{\sqrt{6} - \sqrt{7}}{\sqrt{6} - \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} \times \frac{\sqrt{7} - \sqrt{8}}{\sqrt{7} - \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} \times \frac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}}$

$= \frac{\sqrt{4} - \sqrt{5}}{{(\sqrt{4})}^{2} - {(\sqrt{5})}^{2}} + \frac{\sqrt{5} - \sqrt{6}}{{(\sqrt{5})}^{2} - {(\sqrt{6})}^{2}} + \frac{\sqrt{6} - \sqrt{7}}{{(\sqrt{6})}^{2} - {(\sqrt{7})}^{2}} + \frac{\sqrt{7} - \sqrt{8}}{{(\sqrt{7})}^{2} - {(\sqrt{8})}^{2}} + \frac{\sqrt{8} - \sqrt{9}}{{(\sqrt{8})}^{2} - {(\sqrt{9})}^{2}}$

$= \frac{\sqrt{4} - \sqrt{5}}{4 - 5} + \frac{\sqrt{5} - \sqrt{6}}{5 - 6} + \frac{\sqrt{6} - \sqrt{7}}{6 - 7} + \frac{\sqrt{7} - \sqrt{8}}{7 - 8} + \frac{\sqrt{8} - \sqrt{9}}{8 - 9}$

$= \frac{\sqrt{4} - \sqrt{5}}{- 1} + \frac{\sqrt{5} - \sqrt{6}}{- 1} + \frac{\sqrt{6} - \sqrt{7}}{- 1} + \frac{\sqrt{7} - \sqrt{8}}{- 1} + \frac{\sqrt{8} - \sqrt{9}}{- 1}$

$= - \sqrt{4} + \sqrt{5} - \sqrt{5} + \sqrt{6} - \sqrt{6} + \sqrt{7} - \sqrt{7} + \sqrt{8} - \sqrt{8} + \sqrt{9}$

$= - \sqrt{4} + \sqrt{9}$

$= - 2 + 3$

= $1$

The fifth question of exercise 1.5 ML Aggarwal Class 9 asks us to find the value of two variables by solving the equation containing surds.

5. Given $a$ and $b$ are rational numbers. Find $a$ and $b$ if:

(i) $\frac{3 - \sqrt{5}}{3 + 2 \sqrt{5}} = \frac{- 19}{11} + a \sqrt{5}$

Answer

We have

$\frac{3 - \sqrt{5}}{3 + 2 \sqrt{5}} = \frac{- 19}{11} + a \sqrt{5}$

Rationalising the denominator of LHS by multiplying by $3 - 2 \sqrt{5}$ in numerator and denominator, we get.

$\frac{3 - \sqrt{5}}{3 + 2 \sqrt{5}} \times \frac{3 - 2 \sqrt{5}}{3 - 2 \sqrt{5}} = \frac{- 19}{11} + a \sqrt{5}$

$\Rightarrow \frac{3 (3 - 2 \sqrt{5}) - \sqrt{5} (3 - 2 \sqrt{5})}{3^{2} - {(2 \sqrt{5})}^{2}} = \frac{- 19}{11} + a \sqrt{5}$

$\Rightarrow \frac{9 - 6 \sqrt{5} - 3 \sqrt{5} + 10}{3^{2} - {(2 \sqrt{5})}^{2}} = \frac{- 19}{11} + a \sqrt{5}$

$\Rightarrow \frac{19 - 9 \sqrt{5}}{9 - 20} = \frac{- 19}{11} + a \sqrt{5}$

$\Rightarrow \frac{19 - 9 \sqrt{5}}{- 11} = \frac{- 19}{11} + a \sqrt{5}$

$\Rightarrow - \frac{19}{11} + \frac{9}{11} \sqrt{5} = \frac{- 19}{11} + a \sqrt{5}$

Comparing on both sides, we get

$a = \frac{9}{11}$ .

(ii) $\frac{\sqrt{2} + \sqrt{3}}{3 \sqrt{2} - 2 \sqrt{3}} = a - b \sqrt{6}$

Answer

We have

$\frac{\sqrt{2} + \sqrt{3}}{3 \sqrt{2} - 2 \sqrt{3}} = a - b \sqrt{6}$

Rationalising the denominator of LHS by multiplying by $3 - 2 \sqrt{5}$ in numerator and denominator, we get.

$\frac{\sqrt{2} + \sqrt{3}}{3 \sqrt{2} - 2 \sqrt{3}} \times \frac{3 \sqrt{2} + 2 \sqrt{3}}{3 \sqrt{2} + 2 \sqrt{3}} = a - b \sqrt{6}$

$\Rightarrow \frac{\sqrt{2} (3 \sqrt{2} + 2 \sqrt{3}) + \sqrt{3} (3 \sqrt{2} + 2 \sqrt{3})}{{(3 \sqrt{2})}^{2} - {(2 \sqrt{3})}^{2}} = a - b \sqrt{6}$

$\Rightarrow \frac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6}{{(3 \sqrt{2})}^{2} - {(2 \sqrt{3})}^{2}} = a - b \sqrt{6}$

$\Rightarrow \frac{12 + 5 \sqrt{6}}{18 - 12} = a - b \sqrt{6}$

$\Rightarrow \frac{12 + 5 \sqrt{6}}{6} = a - b \sqrt{6}$

$\Rightarrow 2 + \frac{5}{6} \sqrt{6} = a - b \sqrt{6}$

$\Rightarrow 2 - (- \frac{5}{6}) \sqrt{6} = a - b \sqrt{6}$

Comparing like terms on both sides, we get

$a = 2, b = - \frac{5}{6}$ .

(iii) $\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11} b \sqrt{5}$

Answer

We have

$\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11} b \sqrt{5}$

Rationalising the denominator of LHS by taking L.C.M. of $7 - \sqrt{5}$ and $7 + \sqrt{5}$ , we get

$\frac{{(7 + \sqrt{5})}^{2} - {(7 - \sqrt{5})}^{2}}{(7 - \sqrt{5}) (7 + \sqrt{5})} = a + \frac{7}{11} b \sqrt{5}$

$\Rightarrow \frac{[7^{2} + {(\sqrt{5})}^{2} + 2 \times 7 \times \sqrt{5}] - [7^{2} + {(\sqrt{5})}^{2} - 2 \times 7 \times \sqrt{5}]}{7^{2} - {(\sqrt{5})}^{2}} = a + \frac{7}{11} b \sqrt{5}$

$\Rightarrow \frac{[49 + 5 + 14 \sqrt{5}] - [49 + 5 - 14 \sqrt{5}]}{49 - 5} = a + \frac{7}{11} b \sqrt{5}$

$\Rightarrow \frac{[54 + 14 \sqrt{5}] - [54 - 14 \sqrt{5}]}{44} = a + \frac{7}{11} b \sqrt{5}$

$\Rightarrow \frac{54 + 14 \sqrt{5} - 54 + 14 \sqrt{5}}{44} = a + \frac{7}{11} b \sqrt{5}$

$\Rightarrow \frac{28 \sqrt{5}}{44} = a + \frac{7}{11} b \sqrt{5}$

$\Rightarrow 0 + \frac{7 \sqrt{5}}{11} = a + \frac{7}{11} b \sqrt{5}$

On comparing like terms on both sides, we get

$a = 0, b = 1$ .

The sixth question of exercise 1.5 ML Aggarwal Class 9 asks us to find the values of $p$ and $q$ from a given equation.

6. If $\frac{7 + 3 \sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5}}{3 - \sqrt{5}} = p + q \sqrt{5}$ , find the value of $p$ and $q$ where $p$ and $q$ are rational numbers.

Answer

We have

$\frac{7 + 3 \sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5}}{3 - \sqrt{5}} = p + q \sqrt{5}$

Rationalising the denominator of LHS by taking L.C.M. of $3 - \sqrt{5}$ and $3 + \sqrt{5}$ , we get

$\frac{(7 + 3 \sqrt{5}) (3 - \sqrt{5}) - (7 - 3 \sqrt{5}) (3 + \sqrt{5})}{(3 + \sqrt{5}) (3 - \sqrt{5})} = p + q \sqrt{5}$

$\Rightarrow \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 - 21 - 7 \sqrt{5} + 9 \sqrt{5} + 15}{3^{2} - {(\sqrt{5})}^{2}} = p + q \sqrt{5}$

$\Rightarrow \frac{21 + 2 \sqrt{5} - 36 + 2 \sqrt{5} + 15}{9 - 5} = p + q \sqrt{5}$

$\Rightarrow \frac{36 + 4 \sqrt{5} - 36}{4} = p + q \sqrt{5}$

$\Rightarrow \frac{4 \sqrt{5}}{4} = p + q \sqrt{5}$

$\Rightarrow \sqrt{5} = p + q \sqrt{5}$

On comparing like terms on both sides, we get

$p = 0, q = 1$ .

The seventh question of exercise 1.5 ML Aggarwal Class 9 asks us to rationalise the denominator and hence evaluate by taking value of root 2 and root 3.

7. Rationalise the denominator of the following and hence evaluate by taking $\sqrt{2} = 1.414$ and $\sqrt{3} = 1.732$ , upto three places of decimal:

(i) $\frac{\sqrt{2}}{2 + \sqrt{2}}$

Answer

We have

$\frac{\sqrt{2}}{2 + \sqrt{2}}$

Rationalising the denominator by multiplying the numerator and denominator by ( $2 - \sqrt{2}$ ), we get

$\frac{\sqrt{2}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$ $= \frac{2 \sqrt{2} - 2}{2^{2} - {(\sqrt{2})}^{2}}$ $= \frac{2 (\sqrt{2} - 1)}{4 - 2}$ $= \frac{2 (\sqrt{2} - 1)}{2}$

= $\sqrt{2} - 1$

Now, putting $\sqrt{2} = 1.414$ , we get

= $1.414 - 1$

= $0.414$ .

(ii) $\frac{1}{\sqrt{3} + \sqrt{2}}$

Answer

We have

$\frac{1}{\sqrt{3} + \sqrt{2}}$

Rationalising the denominator by multiplying the numerator and denominator by $(\sqrt{3} - \sqrt{2})$ , we get

$\frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$= \frac{\sqrt{3} - \sqrt{2}}{{(\sqrt{3})}^{2} - {(\sqrt{2})}^{2}}$

$= \frac{\sqrt{3} - \sqrt{2}}{3 - 2}$

$= \sqrt{3} - \sqrt{2}$

Putting $\sqrt{3} = 1.732, \sqrt{2} = 1.414$ , we get

$= 1.732 - 1.414$

= $0.318$ .

The eighth question of exercise 1.5 ML Aggarwal Class 9 asks us to find value of a surd expression.

8. If $a = 2 + \sqrt{3}$ , then find the value of $a - \frac{1}{a}$ .

Answer

We have

$a = 2 + \sqrt{3}$

$∴ \frac{1}{a} = \frac{1}{2 + \sqrt{3}}$ $= \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$

= $\frac{2 - \sqrt{3}}{2^{2} - {(\sqrt{3})}^{2}}$ = $\frac{2 - \sqrt{3}}{4 - 3}$

= $2 - \sqrt{3}$

Now,

$a - \frac{1}{a}$

$= (2 + \sqrt{3}) - (2 - \sqrt{3})$

$= 2 + \sqrt{3} - 2 + \sqrt{3}$

= $2 \sqrt{3}$ .

The ninth question of exercise 1.5 ML Aggarwal Class 9 asks us to find the value of a surd expression.

9. If $x = 1 - \sqrt{2}$ , find the value of ${(x - \frac{1}{x})}^{4}$ .

Answer

We have

$x = 1 - \sqrt{2}$

$\Rightarrow \frac{1}{x} = \frac{1}{1 - \sqrt{2}} \times \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1 + \sqrt{2}}{1^{2} - {(\sqrt{2})}^{2}}$

$= \frac{1 + \sqrt{2}}{1 - 2} = \frac{1 + \sqrt{2}}{- 1}$

$= - 1 - \sqrt{2}$

Now,

$x - \frac{1}{x}$

$= (1 - \sqrt{2}) - (- 1 - \sqrt{2})$

$= 1 - \sqrt{2} + 1 + \sqrt{2}$

= $2$ .

Hence, ${(x - \frac{1}{x})}^{4}$ = ${(2)}^{4}$ = $16$ .

The tenth question of exercise 1.5 ML Aggarwal Class 9 asks us to find the value of an expression.

10. If $x = 5 - 2 \sqrt{6}$ , find the value of $x^{2} + \frac{1}{x^{2}}$ .

Answer

We have

$x = 5 - 2 \sqrt{6}$

$\Rightarrow \frac{1}{x} = \frac{1}{(5 - 2 \sqrt{6})}$ = $\frac{1}{(5 - 2 \sqrt{6})} \times \frac{(5 + 2 \sqrt{6})}{(5 + 2 \sqrt{6})}$

$= \frac{(5 + 2 \sqrt{6})}{5^{2} - {(2 \sqrt{6})}^{2}}$ $= \frac{(5 + 2 \sqrt{6})}{25 - 24}$ $= \frac{(5 + 2 \sqrt{6})}{1}$

$= 5 + 2 \sqrt{6}$

Now,

$x^{2} + \frac{1}{x^{2}} = x^{2} + \frac{1}{x^{2}} + 2 - 2$ $= (x^{2} + \frac{1}{x^{2}} + 2) - 2$

= ${(x + \frac{1}{x})}^{2} - 2$
Putting the value of $x$ and $\frac{1}{x}$ , we get

${(5 - 2 \sqrt{6} + 5 + 2 \sqrt{6})}^{2} - 2$

= ${(5 + 5)}^{2} - 2$

= ${(10)}^{2} - 2$ = $100 - 2$ = $98$ .

The eleventh question of exercise 1.5 ML Aggarwal Class 9 asks us to find the value of an expression.

11. If $p = \frac{2 - \sqrt{5}}{2 + \sqrt{5}}$ and $q = \frac{2 + \sqrt{5}}{2 - \sqrt{5}}$ , find the value of :

(i) $p + q$

Answer

$p + q$

= $\frac{2 - \sqrt{5}}{2 + \sqrt{5}}$ + $\frac{2 + \sqrt{5}}{2 - \sqrt{5}}$

Solving by taking LCM of the denominators

$= \frac{{(2 - \sqrt{5})}^{2} + {(2 + \sqrt{5})}^{2}}{(2 + \sqrt{5}) (2 - \sqrt{5})}$

$= \frac{2^{2} + {(\sqrt{5})}^{2} - 2 \times 2 \times \sqrt{5} + 2^{2} + {(\sqrt{5})}^{2} + 2 \times 2 \times \sqrt{5}}{(2 + \sqrt{5}) (2 - \sqrt{5})}$

$= \frac{4 + 5 - 4 \sqrt{5} + 4 + 5 + 4 \sqrt{5}}{2^{2} - {(\sqrt{5})}^{2}}$

$= \frac{9 + 9}{4 - 5}$

$= \frac{9 + 9}{4 - 5}$ $= \frac{18}{- 1}$ = $- 18$ .

(ii) $p - q$

Answer

$p - q$

= $\frac{2 - \sqrt{5}}{2 + \sqrt{5}}$ – $\frac{2 + \sqrt{5}}{2 - \sqrt{5}}$

Solving by taking LCM of the denominators

$= \frac{{(2 - \sqrt{5})}^{2} - {(2 + \sqrt{5})}^{2}}{(2 + \sqrt{5}) (2 - \sqrt{5})}$

$= \frac{2^{2} + {(\sqrt{5})}^{2} - 2 \times 2 \times \sqrt{5} - 2^{2} - {(\sqrt{5})}^{2} - 2 \times 2 \times \sqrt{5}}{(2 + \sqrt{5}) (2 - \sqrt{5})}$

$= \frac{4 + 5 - 4 \sqrt{5} - 4 - 5 - 4 \sqrt{5}}{2^{2} - {(\sqrt{5})}^{2}}$

$= \frac{- 4 \sqrt{5} - 4 \sqrt{5}}{- 1}$

$= \frac{- 8 \sqrt{5}}{- 1}$

= $8 \sqrt{5}$

(iii) $p^{2} + q^{2}$

Answer

$p^{2} + q^{2}$

$= p^{2} + q^{2} + 2 p q - 2 p q$

$= {(p + q)}^{2} - 2 p q$

From (i) $p + q = - 18$

$= {(- 18)}^{2} - 2 \times \frac{2 - \sqrt{5}}{2 + \sqrt{5}} \times \frac{2 + \sqrt{5}}{2 - \sqrt{5}}$

$= 324 - 2 \times 1$

$= 324 - 2$ = $322$ .

(iv) $p^{2} - q^{2}$

Answer

$p^{2} - q^{2}$

= $(p + q) (p - q)$

From (i) and (ii), we have
$(p + q) = - 18, (p - q) = 8 \sqrt{5}$

= $- 18 \times 8 \sqrt{5}$ = $- 144 \sqrt{5}$ .

The eleventh question of exercise 1.5 ML Aggarwal Class 9 asks us to find the value of an expression.

12. If $x = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$ and $y = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$ , then find the value of $x^{2} + 5 x y + y^{2}$ .

Answer

We have $x = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$ and $y = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$

$∴ x^{2} + 5 x y + y^{2}$

$= (x^{2} + 2 x y + y^{2}) + 3 x y$

$= {(x + y)}^{2} + 3 x y$

$= {(\frac{\sqrt{2} - 1}{\sqrt{2} + 1} + \frac{\sqrt{2} + 1}{\sqrt{2} - 1})}^{2} + 3 \times \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$

$= {(\frac{{(\sqrt{2} - 1)}^{2} + {(\sqrt{2} + 1)}^{2}}{(\sqrt{2} + 1) (\sqrt{2} - 1)})}^{2} + 3 \times 1$

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

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

$= {(\frac{3 + 3}{1})}^{2} + 3$

$= {(6)}^{2} + 3$

$= 36 + 3$ = $39$ .

Frequently Asked Questions

How to rationalise the denominator of $. \frac{1}{a + \sqrt{b}}$

To rationalise the denominator of $\frac{1}{a + \sqrt{b}}$ , we multiply by the conjugate of $(a + \sqrt{b})$ . The conjugate of $(a + \sqrt{b})$ is $(a - \sqrt{b})$ . So, we multiply by $(a - \sqrt{b})$ in the numerator and denominator both to make the denominator rational.

$\frac{1}{(a - \sqrt{b})} \times \frac{(a + \sqrt{b})}{(a + \sqrt{b})}$ = $\frac{(a + \sqrt{b})}{a^{2} - {(\sqrt{b})}^{2}}$ = $\frac{a + \sqrt{b}}{a^{2} - b}$

What is rationalisation?

The process of multiplying a surd by another surd to get a rational number is called rationalisation. For example, to rationalise $\frac{1}{\sqrt{2}}$ , we multiply by $\sqrt{2}$ in numerator and denominator both so that denominator becomes rational.

Why do we rationalise the denominator of a surd?

We rationalise the denominator of a surd because it is difficult to simplify an irrational number. So, first we make it rational and then simplification becomes simple as we can operate usual additions, subtractions, multiplications and divisions.

Exercise 1.5 ML Aggarwal Class 9 Mathematics Solutions

1. Rationalise the denominator of the following:

(i) 345

(ii) 573

(iii) 34−7

(iv) 1732+1

=1732+1×32−132−1

(v) 1641−5

(vi) 17−6

(vii) 15+2

(viii) 2+32−3

2. Simplify each of the following by rationalising the denominator:

(i) 7+357−35

(ii) 3−223+22

(iii) 5−3147+214

3. Simplify: 7310+3−256+5−3215+32.

4. Simplify: 14+5+15+6+16+7+17+8+18+9.

5. Given a and b are rational numbers. Find a and b if:

(i) 3−53+25=−1911+a5

(ii) 2+332−23=a−b6

(iii) 7+57−5−7−57+5=a+711b5

6. If 7+353+5−7−353−5=p+q5, find the value of p and q where p and q are rational numbers.

7. Rationalise the denominator of the following and hence evaluate by taking 2=1.414 and 3=1.732, upto three places of decimal:

(i) 22+2

(ii) 13+2

8. If a=2+3, then find the value of a−1a.

9. If x=1−2, find the value of (x−1x)4.

10. If x=5−26, find the value of x2+1x2.

11. If p=2−52+5 and q=2+52−5, find the value of :

(i) p+q

(ii) p−q

(iii) p2+q2

(iv) p2−q2

12. If x=2−12+1 and y=2+12−1, then find the value of x2+5xy+y2.

Frequently Asked Questions

How to rationalise the denominator of .1a+b

What is rationalisation?

Why do we rationalise the denominator of a surd?

(i) $\frac{3}{4 \sqrt{5}}$

(ii) $\frac{5 \sqrt{7}}{\sqrt{3}}$

(iii) $\frac{3}{4 - \sqrt{7}}$

(iv) $\frac{17}{3 \sqrt{2} + 1}$

$= \frac{17}{3 \sqrt{2} + 1} \times \frac{3 \sqrt{2} - 1}{3 \sqrt{2} - 1}$

(v) $\frac{16}{\sqrt{41} - 5}$

(vi) $\frac{1}{\sqrt{7} - \sqrt{6}}$

(vii) $\frac{1}{\sqrt{5} + \sqrt{2}}$

(viii) $\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}$

(i) $\frac{7 + 3 \sqrt{5}}{7 - 3 \sqrt{5}}$

(ii) $\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}}$

(iii) $\frac{5 - 3 \sqrt{14}}{7 + 2 \sqrt{14}}$

3. Simplify: $\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3 \sqrt{2}}{\sqrt{15} + 3 \sqrt{2}}$ .

4. Simplify: $\frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}$ .

5. Given $a$ and $b$ are rational numbers. Find $a$ and $b$ if:

(i) $\frac{3 - \sqrt{5}}{3 + 2 \sqrt{5}} = \frac{- 19}{11} + a \sqrt{5}$

(ii) $\frac{\sqrt{2} + \sqrt{3}}{3 \sqrt{2} - 2 \sqrt{3}} = a - b \sqrt{6}$

(iii) $\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11} b \sqrt{5}$

6. If $\frac{7 + 3 \sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5}}{3 - \sqrt{5}} = p + q \sqrt{5}$ , find the value of $p$ and $q$ where $p$ and $q$ are rational numbers.

7. Rationalise the denominator of the following and hence evaluate by taking $\sqrt{2} = 1.414$ and $\sqrt{3} = 1.732$ , upto three places of decimal:

(i) $\frac{\sqrt{2}}{2 + \sqrt{2}}$

(ii) $\frac{1}{\sqrt{3} + \sqrt{2}}$

8. If $a = 2 + \sqrt{3}$ , then find the value of $a - \frac{1}{a}$ .

9. If $x = 1 - \sqrt{2}$ , find the value of ${(x - \frac{1}{x})}^{4}$ .

10. If $x = 5 - 2 \sqrt{6}$ , find the value of $x^{2} + \frac{1}{x^{2}}$ .

11. If $p = \frac{2 - \sqrt{5}}{2 + \sqrt{5}}$ and $q = \frac{2 + \sqrt{5}}{2 - \sqrt{5}}$ , find the value of :

(i) $p + q$

(ii) $p - q$

(iii) $p^{2} + q^{2}$

(iv) $p^{2} - q^{2}$

12. If $x = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$ and $y = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$ , then find the value of $x^{2} + 5 x y + y^{2}$ .

How to rationalise the denominator of $. \frac{1}{a + \sqrt{b}}$