Selina Solutions Class 9 Mathematics Solutions for Exercise 1(C) in Chapter 1 - Chapter 1- Rational and Irrational Numbers

Question 13 Exercise 1(C)

If √𝟐 = 𝟏. 𝟒 and √𝟑 = 𝟏. 𝟕, find the value of each of the following, correct to

one decimal place:

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

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

(iii) $\frac{2-\sqrt{3}}{\sqrt{3}}$

Answer:

$\begin{aligned} &\text { (i) }\\ &\begin{array}{l} \sqrt{2}=1.4 \text { and } \sqrt{3}=1.7 \\ \frac{1}{\sqrt{3}-\sqrt{2}} \\ =\frac{1}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}} \\ =\frac{\sqrt{3}+\sqrt{2}}{3-2} \\ =\sqrt{3}+\sqrt{2} \\ =1.7+1.4 \\ =3.1 \end{array} \end{aligned}$

$(ii) $\sqrt{2}=1.4$ and $\sqrt{3}=1.7$ (ii) $\frac{1}{3+2 \sqrt{2}}$ $=\frac{1}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}$ $=\frac{3-2 \sqrt{2}}{(3)^{2}-(2 \sqrt{2})^{2}}$ $=\frac{3-2 \sqrt{2}}{9-8}$ $=3-2 \sqrt{2}$ $=3-2(1.4)$ $=3-2.8$ $=0.2$$

$\begin{aligned} &\text { (iii) }\\ &\sqrt{2}=1.4 \text { and } \sqrt{3}=1.7 \end{aligned}$

$\begin{array}{l} \frac{2-\sqrt{3}}{\frac{2-\sqrt{3}}{\sqrt{3}}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \frac{2 \sqrt{3}-3}{3 \cdot 4-3}=\frac{(2 \times 1.7)-3}{3} \\ \frac{3}{3}=\frac{0.4}{3}=0.1 \end{array}$

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