Students often turn to OP Malhotra Class 9 Solutions Chapter 1 Rational and Irrational Numbers Ex 1(C) to clarify doubts and improve problem-solving skills.

## S Chand Class 9 ICSE Maths Solutions Chapter 1 Rational and Irrational Numbers Ex 1(C)

Simplify:

Question 1.
(a) $$\sqrt{\frac{1}{3}}$$
(b) $$\sqrt{\frac{5}{12}}$$
(c) $$\sqrt{1 \frac{46}{75}}$$
Solution:

Question 2.
$$\sqrt{112}-\sqrt{63}+\frac{224}{\sqrt{28}}$$
Solution:

Question 3.
$$\frac{4 \sqrt{18}}{\sqrt{12}}-\frac{8 \sqrt{75}}{\sqrt{32}}+\frac{9 \sqrt{2}}{\sqrt{3}}$$
Solution:

Question 4.
Rationalise the denominators of
(a) $$\frac{1}{4-\sqrt{3}}$$
(b) $$\frac{2}{\sqrt{5}+\sqrt{3}}$$
(c) $$\frac{1}{2 \sqrt{5}-\sqrt{3}}$$
(d) $$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$
Solution:

Question 5.
Rationalise the denominator and simplify :
(i) $$\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$$
(ii) $$\frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}}$$
Solution:

Question 6.
Find the values of a and b if
(i) $$\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2}$$
(ii) $$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}$$
(iii) $$\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b \sqrt{7}$$
Solution:

Question 7.
Rationalise the denominator of $$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$.
Solution:
$$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{5}}$$
Rationalising denominator
= $$\frac{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}{[(\sqrt{2}+\sqrt{3})+\sqrt{5}][(\sqrt{2}+\sqrt{3})-\sqrt{5}]}$$
= $$\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2 \sqrt{6}-5}$$
= $$\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2 \sqrt{6}}$$
Again rationalising,
= $$\frac{(\sqrt{2}+\sqrt{3}-\sqrt{5}) \sqrt{6}}{2 \times \sqrt{6} \times \sqrt{6}}=\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{2 \times 6}$$
= $$\frac{\sqrt{4 \times 3}+\sqrt{9 \times 2}-\sqrt{30}}{12}$$
= $$\frac{2 \sqrt{3}+3 \sqrt{2}-\sqrt{30}}{12}$$

Question 8.
Taking $$\sqrt{2}$$ = 1.414 and $$\sqrt{3}$$ = 1.732, find without using tables or long division, the value of
(a) $$\frac{1}{3-\sqrt{2}}$$
(b) $$\frac{2}{\sqrt{3}-\sqrt{2}}$$
Solution:
(a) $$\frac{1}{3-\sqrt{2}}=\frac{1(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$$
(Rationalising the denominator)
= $$\frac{3+\sqrt{2}}{(3)^2-(\sqrt{2})^2}$$
= $$\frac{3+\sqrt{2}}{9-2}=\frac{3+\sqrt{2}}{7}=\frac{3+1.414}{7}$$
= $$\frac { 4.414 }{ 7 }$$ = 0.631

(b) $$\frac{2}{\sqrt{3}-\sqrt{2}}=\frac{2(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$$
(Rationalising the denominator)
= $$\frac{2(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{2(\sqrt{3}+\sqrt{2})}{3-2}$$
= 2($$\sqrt{3}$$ + $$\sqrt{2}$$)
= 2 (1.732 + 1.414) = 2 x 3.146
= 6.292

Question 9.
Express $$\frac{3-5 \sqrt{5}}{3+2 \sqrt{5}}$$ in the form (a$$\sqrt{5}$$ – b) where a and b are simple fractions.
Solution:

Question 10.
Prove that $$\frac{1}{\sqrt{2}-1}+\frac{2}{\sqrt{3}+1}=\sqrt{2}+\sqrt{3}$$.
Solution:

Question 11.
Simplify:
$$\frac{6 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$$ (NDA)
Solution:

Question 12.
Simplify:
(i) $$\frac{6}{2 \sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}$$
(ii) $$\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$$
Solution:
(i) $$\frac{6}{2 \sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}$$

Now from (i), (ii), (iii)
(2$$\sqrt{3}$$ + $$\sqrt{6}$$) + (3$$\sqrt{2}$$ – 2$$\sqrt{3}$$) – (3$$\sqrt{2}$$ + $$\sqrt{6}$$)
= 2$$\sqrt{3}$$ + $$\sqrt{6}$$ + 3$$\sqrt{2}$$ – 2$$\sqrt{2}$$ – 3$$\sqrt{2}$$ – $$\sqrt{6}$$ = 0

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

From (i), (ii) and (iii)
($$\sqrt{30}$$ – 3) – (2$$\sqrt{30}$$ – 10) – (- $$\sqrt{30}$$ + 6)
= $$\sqrt{30}$$ – 3 – 27$$\sqrt{30}$$ + 10 + $$\sqrt{30}$$ – 6
= 10 – 9 = 1

Question 13.
If x = 2 + $$\sqrt{3}$$, find the value of x² + $$\frac{1}{x^2}$$.
Solution:
x = 2 + $$\sqrt{3}$$
$$\frac { 1 }{ x }$$ = $$\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}$$
= $$\frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}$$
Now x + $$\frac { 1 }{ x }$$ = 2 + $$\sqrt{3}$$ + 2 – $$\sqrt{3}$$ = 4
Squaring both sides,
⇒ $$\left(x+\frac{1}{x}\right)^2$$ = (4)²
⇒ x² + $$\frac { 1 }{ x² }$$ + 2 = 16
⇒ x² + $$\frac { 1 }{ x² }$$ = 16 – 2 = 14

Question 14.
If x = $$\sqrt{2}$$ + 1, find the value of x² + $$\frac{1}{x^2}$$.
Solution:
x = $$\sqrt{2}$$ + 1
∴ $$\frac { 1 }{ x }$$ = $$\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}$$
= $$\frac{\sqrt{2}-1}{2-1}=\sqrt{2}$$ – 1
Now x + $$\frac { 1 }{ x }$$ = $$\sqrt{2}$$ + 1 + $$\sqrt{2}$$ – 1
Squaring both sides,
⇒ $$\left(x+\frac{1}{x}\right)^2=(2 \sqrt{2})^2$$
⇒ x² + $$\frac { 1 }{ x² }$$ + 2 = 4 x 2 = 8
⇒ x² + $$\frac { 1 }{ x² }$$ = 8 – 2 = 6

Question 15.
If x = $$\frac{5-\sqrt{21}}{2}$$, find the value of
(i) x + $$\frac { 1 }{ x }$$ and (ii) x² + $$\frac { 1 }{ x² }$$
Solution:
x = $$\frac{5-\sqrt{21}}{2}$$
∴ $$\frac { 1 }{ x }$$ = $$\frac{2}{5-\sqrt{21}}=\frac{2(5+\sqrt{21})}{(5-\sqrt{21})(5+\sqrt{21})}$$
= $$\frac{2(5+\sqrt{21})}{25-21}=\frac{2(5+\sqrt{21})}{4}=\frac{5+\sqrt{21}}{2}$$

(i) x + $$\frac { 1 }{ x }$$ = $$\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}$$
= $$\frac{5-\sqrt{21}+5+\sqrt{21}}{2}$$
= $$\frac { 10 }{ 2 }$$ = 5
and 1 + $$\frac { 1 }{ x }$$ = 5
Squaring both sides,
(x + $$\frac { 1 }{ x }$$)² = (5)²
⇒ x² + $$\frac { 1 }{ x² }$$ + 2 = 25
⇒ x² + $$\frac { 1 }{ x² }$$ = 25 – 2 = 23
Hence x + $$\frac { 1 }{ x }$$ = 5 and x² + $$\frac { 1 }{ x² }$$ = 23