Students can download Maths Chapter 2 Real Numbers Ex 2.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.7

Question 1.
Rationalise the denominator:
(i) $$\frac{1}{\sqrt{50}}$$
(ii) $$\frac{5}{3\sqrt{5}}$$
(iii) $$\frac{\sqrt{75}}{\sqrt{18}}$$
(iv) $$\frac{3\sqrt{5}}{\sqrt{6}}$$
Solution:
(i) $$\frac{1}{\sqrt{50}}$$ = $$\frac{1}{\sqrt{25 \times 2}}=\frac{1}{5 \sqrt{2}}=\frac{1}{5 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{5 \times 2}=\frac{\sqrt{2}}{10}$$

(ii) $$\frac{5}{3\sqrt{5}}$$ = $$\frac{5}{3 \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{5 \sqrt{5}}{3 \times 5}=\frac{\sqrt{5}}{3}$$

(iii) $$\frac{\sqrt{75}}{\sqrt{18}}$$ = $$\frac{\sqrt{3 \times 25}}{\sqrt{2 \times 9}}=\frac{5 \sqrt{3}}{3 \sqrt{2}}=\frac{5 \sqrt{3}}{3 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{5 \sqrt{6}}{3 \times 2}=\frac{5 \sqrt{6}}{6}$$

(iv) $$\frac{3\sqrt{5}}{\sqrt{6}}$$ = $$\frac{3 \sqrt{5}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=\frac{3 \sqrt{30}}{6}=\frac{\sqrt{30}}{2}$$

Question 2.
Rationalise the denominator and simplify:
(i) $$\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}$$
Solution:

(ii) $$\frac{5\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$
Solution:

(iii) $$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$$
Solution:

(iv) $$\frac{\sqrt{5}}{\sqrt{6}+2} – \frac{\sqrt{5}}{\sqrt{6}-2}$$
Solution:

Question 3.
Find the value of a and b if $$\frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b$$.
Solution:

Question 4.
If x = $$\sqrt{7}$$ + 2, then find the value of x² + $$\frac{1}{x^2}$$
Solution:
$$\sqrt{7}$$ + 2 ⇒ x² = $$(\sqrt{5}+2)^{2}$$
= $$(\sqrt{5})^{2}$$ + 2 × 2 × $$\sqrt{5}$$ + 2² = 5 + 4 $$\sqrt{5}$$ + 4 = 9 + 4$$\sqrt{5}$$
$$\frac{1}{x}=\frac{1}{\sqrt{5}+2}=\frac{\sqrt{5}-2}{(\sqrt{5}+2)(\sqrt{5}-2)}=\frac{\sqrt{5}-2}{(\sqrt{5})^{2}-2^{2}}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2$$
$$\frac{1}{x^{2}}$$ = ($$\sqrt{5} – 2)^{2}$$
= $$(\sqrt{5})^{2}$$ – 2 × $$\sqrt{5}$$ × 2 + 2² = 5 – 4 $$\sqrt{5}$$ + 4 = 9 – 4 $$\sqrt{5}$$
∴ x² + $$\frac{1}{x^{2}}$$ = 9 + $$4\sqrt{5}$$ + 9 – $$4\sqrt{5}$$ = 18
The value of x² + $$\frac{1}{x^{2}}$$ = 18

Question 5.
Given $$\sqrt{2}$$ = 1.414, find the value of $$\frac{8 – 5\sqrt{2}}{3 – 2\sqrt{2}}$$ (to 3 places of decimals).
Solution:

= 4 + $$\sqrt{2}$$ = 4 + 1.414 = 5.414