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

Question 4 Exercise 1(C)

Find the values of ‘a’ and ‘b’ in each of the following:

\begin{aligned} &(i)\\ &\frac{2+\sqrt{3}}{2-\sqrt{3}}=a+b \sqrt{3} \end{aligned}

\text { (ii) } \quad \frac{\sqrt{7}-2}{\sqrt{7}+2}=a \sqrt{7}+b

\text { (iii) } \frac{3}{\sqrt{3}-\sqrt{2}}=a \sqrt{3}+b \sqrt{2}

\text { (iv) } \frac{5+3 \sqrt{2}}{5-3 \sqrt{2}}=a+b \sqrt{2}

Answer:

(i) \frac{2+\sqrt{3}}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}=a+b \sqrt{3}

\begin{array}{l} \frac{(2+\sqrt{3})^{2}}{(2)^{2}-(\sqrt{3})^{2}}=a+b \sqrt{3} \\ \frac{4+3+4 \sqrt{3}}{4-3}=a+b \sqrt{3} \\ 7+4 \sqrt{3}=a+b \sqrt{3} \\ a=7, b=4 \end{array}

(ii) \begin{array}{l} \frac{\sqrt{7}-2}{\sqrt{7}+2} \times \frac{\sqrt{7}-2}{\sqrt{7}-2}=a \sqrt{7}+b \\ \frac{(\sqrt{7}-2)^{2}}{(\sqrt{7})^{2}-(2)^{2}}=a \sqrt{7}+b \\ \frac{7+4-4 \sqrt{7}}{7-4}=a \sqrt{7}+b \\ \frac{11-4 \sqrt{7}}{3}=8 \sqrt{7}+b \\ a=\frac{-4}{3}, b=\frac{11}{3} \end{array}

(iii) \begin{array}{l} \sqrt{3}-\sqrt{2} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=a \sqrt{3}-b \sqrt{2} \\ \frac{3(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=2 \sqrt{3}-b \sqrt{2} \\ \frac{3(\sqrt{3}+\sqrt{2})}{3-2}=a \sqrt{3}-b \sqrt{2} \\ (3 \sqrt{3}+3 \sqrt{2})=a \sqrt{3}-b \sqrt{2} \end{array}

(iv)\frac{5+3 \sqrt{2}}{5-3 \sqrt{2}} \times \frac{5+3 \sqrt{2}}{5+3 \sqrt{2}}=a+b \sqrt{2}'

\begin{array}{l} \frac{(5+3 \sqrt{2})^{2}}{(5)^{2}-(3 \sqrt{2})^{2}}=a+b \sqrt{2} \\ \frac{25+18+30 \sqrt{2}}{25-18}=a+b \sqrt{2} \\ \frac{43+30 \sqrt{2}}{7}=a+b \sqrt{2} \\ a=\frac{43}{7}, \quad b=\frac{30}{7} \end{array}

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