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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.6

Question 1.
Simplify the following using addition and subtraction properties of surds:
(i) 5$$\sqrt{3}$$ + 18$$\sqrt{3}$$ – 2$$\sqrt{3}$$
(ii) 4$$\sqrt{5}$$ + 2$$\sqrt{5}$$ – 3$$\sqrt{5}$$
(iii) 3$$\sqrt{75}$$ + 5$$\sqrt{48}$$ – $$\sqrt{243}$$
(iv) 5$$\sqrt{40}$$ + 2$$\sqrt{625}$$ – 3$$\sqrt{320}$$
Solution:
(i) 5$$\sqrt{3}$$ + 18$$\sqrt{3}$$ – 2$$\sqrt{3}$$ = (5 + 18 – 2)$$\sqrt{3}$$
= (23 – 2) $$\sqrt{3}$$ = 21$$\sqrt{3}$$

(ii) 4$$\sqrt{5}$$ + 2$$\sqrt{5}$$ – 3$$\sqrt{5}$$ = (4 + 2 – 3) $$\sqrt{5}$$
= (6 – 3) $$\sqrt{5}$$ = 3$$\sqrt{5}$$ (iii) 3$$\sqrt{75}$$ + 5$$\sqrt{48}$$ – $$\sqrt{243}$$ = $$3\sqrt{5^{2}×3} + 5\sqrt{2^{4}×3} – \sqrt{3^{5}}$$ = 3 × 5$$\sqrt{3}$$ + 5 × 2²$$\sqrt{3}$$ – 3²$$\sqrt{3}$$ = 15$$\sqrt{3}$$ + 20$$\sqrt{3}$$ – 9$$\sqrt{3}$$
= (15 + 20 – 9)$$\sqrt{3}$$
= (35 – 9)$$\sqrt{3}$$
= 26 $$\sqrt{3}$$

(iv) 5$$\sqrt{40}$$ + 2$$\sqrt{625}$$ – 3$$\sqrt{320}$$ = 5$$\sqrt{2^{3}×5} + 2\sqrt{5^{3}×5} – 3\sqrt{2^{3}×2^{3}×5}$$
5 × 2$$\sqrt{5} + 2 × 5\sqrt{5} – 3 × 2 × 2\sqrt{5}$$
= 10$$\sqrt{5} + 10\sqrt{5} – 12\sqrt{5}$$
= 20$$\sqrt{5} – 12\sqrt{5}$$
= 8$$\sqrt{5}$$ Question 2.
Simplify the following using multiplication and division properties of surds:
(i) $$\sqrt{3}$$ × $$\sqrt{5}$$ × $$\sqrt{2}$$
(ii) $$\sqrt{35}$$ ÷ $$\sqrt{7}$$
(iii) $$\sqrt{27}$$ × $$\sqrt{8}$$ × $$\sqrt{125}$$
(iv) (7$$\sqrt{a}$$ – 5$$\sqrt{b}$$) (7$$\sqrt{a}$$ + 5$$\sqrt{b}$$)
(v) ($$\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}$$) ÷ $$\sqrt{\frac{16}{81}}$$
Solution:
(i) $$\sqrt{3}$$ × $$\sqrt{5}$$ × $$\sqrt{2}$$ = $$\sqrt{3×5×2} = \sqrt{30}$$

(ii) $$\sqrt{37} ÷ \sqrt{7} = \frac{\sqrt{35}}{\sqrt{7}} = \sqrt{\frac{35}{7}} = \sqrt{5}$$

(iii) $$\sqrt{27}$$ × $$\sqrt{8}$$ × $$\sqrt{125}$$ = $$\sqrt{27×8×125}$$
= $$\sqrt{3^{3}×2^{3}×5^{3}}$$ = 3 × 2 × 5 = 30

(iv) (7$$\sqrt{a}$$ – 5$$\sqrt{b}$$) (7$$\sqrt{a}$$ + 5$$\sqrt{b}$$)
[using a2 – b2 = (a + b) (a – b)]
(7$$\sqrt{a}$$ – 5$$\sqrt{b}$$) (7$$\sqrt{a}$$ + 5$$\sqrt{b}$$) = $$(7\sqrt{a})^{2} – (5\sqrt{b})^{2}$$ = 49a – 25b

(v) ($$\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}$$) ÷ $$\sqrt{\frac{16}{81}}$$  = $$\frac{5}{36}$$ × $$\frac{9}{4}$$
= $$\frac{5×1}{4×4}$$
= $$\frac{5}{16}$$ Question 3.
If $$\sqrt{2}$$ = 1.414, $$\sqrt{3}$$ = 1.732, $$\sqrt{5}$$ = 2.236, $$\sqrt{10}$$ = 3.162, then find the values of the following correct to 3 places of decimals.
(i) $$\sqrt{40}$$ – $$\sqrt{20}$$
(ii) $$\sqrt{300}$$ + $$\sqrt{90}$$ – $$\sqrt{8}$$
Solution:
(i) $$\sqrt{40}$$ – $$\sqrt{20}$$ = $$\sqrt{4×10} – \sqrt{4×5} = 2\sqrt{10} – 2\sqrt{5}$$
= 2 × 3.162 – 2 × 2.236 = 6.324 – 4.472 = 1.852

(ii) $$\sqrt{300}$$ + $$\sqrt{90}$$ – $$\sqrt{8}$$ = $$\sqrt{3×100} + \sqrt{9×10} – \sqrt{4×2}$$
= 10$$\sqrt{3}$$ + 3$$\sqrt{10}$$ – 2$$\sqrt{2}$$
= 10 × 1.732 + 3 × 3.162 – 2 × 1.414
= 17.32 + 9.486 – 2.828
= 26.806 – 2.828
= 23.978 Question 4.
Arrange surds in descending order
(i) $$\sqrt{5}$$, $$\sqrt{4}$$, $$\sqrt{3}$$
Solution:
LCM of 3, 9 and 6 is 18 $$\sqrt{5}$$ = $$\sqrt[3×6]{5^{6}}$$ = $$\sqrt{15625}$$
$$\sqrt{4}$$ = $$\sqrt[2×9]{4^{2}}$$ = $$\sqrt{16}$$
$$\sqrt{3}$$ = $$\sqrt[3×6]{3^{3}}$$ = $$\sqrt{27}$$
$$\sqrt{15625}$$ > $$\sqrt{27}$$ > $$\sqrt{16}$$
$$\sqrt{5}$$ > $$\sqrt{3}$$ > $$\sqrt{4}$$ (ii) $$\sqrt{\sqrt{5}}$$, $$\sqrt{\sqrt{7}}$$, $$\sqrt{\sqrt{3}}$$
Solution:
$$\sqrt{\sqrt{5}}$$ = $$\sqrt{5}$$; $$\sqrt{\sqrt{7}}$$ = $$\sqrt{7}$$; $$\sqrt{\sqrt{3}}$$ = $$\sqrt{3}$$
LCM of 6, 12 and 4 is 12 $$\sqrt{\sqrt{5}}$$ = $$\sqrt{5}$$ = $$\sqrt{5^{2}}$$ = $$\sqrt{25}$$
$$\sqrt{\sqrt{7}}$$ = $$\sqrt{7}$$ = $$\sqrt{7}$$
$$\sqrt{\sqrt{3}}$$ = $$\sqrt{3}$$ = $$\sqrt{3^{3}}$$ = $$\sqrt{27}$$
$$\sqrt{27}$$ > $$\sqrt{25}$$ > $$\sqrt{7}$$
$$\sqrt{\sqrt{3}}$$ > $$\sqrt{\sqrt{5}}$$ > $$\sqrt{\sqrt{7}}$$

Question 5.
Can you get a pure surd when you find:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes we can get a surd.
Example:
(a) 3$$\sqrt{2}$$ + 5$$\sqrt{2}$$ = (3 + 5)$$\sqrt{2}$$ = 8$$\sqrt{2}$$
(b) 3$$\sqrt{6}$$ + 2$$\sqrt{6}$$ = (3 + 2)$$\sqrt{6}$$ = 5$$\sqrt{6}$$

(ii) Yes we can get a surd.
Example:
(a) $$\sqrt{75}$$ – $$\sqrt{48}$$ = $$\sqrt{25×3}$$ – $$\sqrt{16×3}$$ = (5 – 4) $$\sqrt{3}$$ = $$\sqrt{3}$$
(b) $$\sqrt{98}$$ – $$\sqrt{72}$$ = $$\sqrt{49×2}$$ – $$\sqrt{36×2}$$ = (7 – 6) $$\sqrt{2}$$ = $$\sqrt{2}$$ (iii) Yes we can get a surd.
Example:
(a) $$\sqrt{8}$$ × $$\sqrt{6}$$ = $$\sqrt{8×6}$$ = $$\sqrt{48}$$
(b) $$\sqrt{11}$$ × $$\sqrt{3}$$ = $$\sqrt{11×3}$$ = $$\sqrt{33}$$

(iv) Yes we can get a surd.
Example:
(a) $$\sqrt{55}$$ ÷ $$\sqrt{5}$$ = $$\frac{\sqrt{11×5}}{\sqrt{5}} = \sqrt{11}$$
(b) $$\sqrt{65}$$ ÷ $$\sqrt{5}$$ = $$\frac{\sqrt{13×5}}{\sqrt{13}} = \sqrt{5}$$

Question 6.
Can you get a rational number when you compute:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes, the sum of two surds will give a rational number.
Example:
(a) (2 + $$\sqrt{3}$$) + (2 – $$\sqrt{3}$$) = 4
(b) ($$\sqrt{5}$$ + 4) + (7 – $$\sqrt{5}$$) = 11

(ii) Yes, the difference of two surds will give a rational number.
Example:
(a) (5 + $$\sqrt{7}$$) – (- 5 + $$\sqrt{7}$$) = 10
(b) ($$\sqrt{11}$$ + 5) – (-3 + $$\sqrt{11}$$) = 8 (iii) Yes, the product of two surds will give a rational number.
Example:
(a) $$\sqrt{125}$$ × $$\sqrt{45}$$ = $$\sqrt{25×5}$$ × $$\sqrt{9×5}$$ = 5$$\sqrt{5}$$ × 3$$\sqrt{5}$$ = 15 × 5 = 75
(b) $$\sqrt{150}$$ × $$\sqrt{6}$$ = $$\sqrt{25×6}$$ × $$\sqrt{6}$$ = 5$$\sqrt{6}$$ × $$\sqrt{6}$$ = 5 × 6 = 30

(iv) Yes. The quotient of two surds will give a rational number.
Example:
(a) $$\sqrt{32}$$ ÷ $$\sqrt{8}$$ = $$\frac{\sqrt{8×4}}{\sqrt{8}} = \sqrt{4}$$ = 2
(b) $$\sqrt{50}$$ ÷ $$\sqrt{2}$$ = $$\frac{\sqrt{25×2}}{\sqrt{2}} = \sqrt{25}$$ = 5