Exercise 2.7 - Chapter 2 Real Numbers 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E} \times 2.7$
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) $\frac{1}{\sqrt{50}}=\frac{1}{\sqrt{50}} \times \frac{\sqrt{50}}{\sqrt{50}}=\frac{\sqrt{50}}{50}=\frac{\sqrt{5 \times 5 \times 2}}{5 \times 5 \times 2}=\frac{\not 5 \sqrt{2}}{\not 5 \times 5 \times 2}=\frac{\sqrt{2}}{10}$
(ii) $\frac{5}{3 \sqrt{5}}=\frac{5}{3 \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\not 5 \sqrt{5}}{3 \times \not}=\frac{\sqrt{5}}{3}$
(iii) $\frac{\sqrt{75}}{\sqrt{18}}=\frac{\sqrt{3 \times 5 \times 5}}{\sqrt{3 \times 2 \times 3}}=\frac{5 \sqrt{3} \times \sqrt{2}}{3 \sqrt{2}} \times \sqrt{2}=\frac{5 \sqrt{6}}{3 \times 2}=\frac{5 \sqrt{6}}{6}$
(iv) $\frac{3 \sqrt{5}}{\sqrt{6}}=\frac{3 \sqrt{5}}{\sqrt{6}} \times \sqrt{6}=\frac{\not j \sqrt{30}}{6}=\frac{\sqrt{30}}{2}$

Question 2.
Rationalise the denominator and simplify
(i) $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}$
(ii) $\frac{5 \sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
(iii) $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$
(iv) $\frac{\sqrt{5}}{\sqrt{6}+2}-\frac{\sqrt{5}}{\sqrt{6}-2}$
Solution:
(i) $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}=\frac{\sqrt{48}+\sqrt{32}}{(\sqrt{27}-\sqrt{18})} \times \frac{\sqrt{27}+\sqrt{18}}{\sqrt{27}+\sqrt{18}}$
Multiply the numerator and denominator by the rationatising factor $(\sqrt{27}+\sqrt{18})$
$=\frac{\sqrt{48 \times 27}+\sqrt{32} \times \sqrt{27}+\sqrt{48 \times 18}+\sqrt{32 \times 18}}{\sqrt{27^{2}}-\sqrt{18^{2}}}$

$=\frac{\sqrt{3 \times 16 \times 3 \times 9}+\sqrt{2 \times 16 \times 3 \times 9}+\sqrt{3 \times 16 \times 2 \times 9}+\sqrt{2 \times 16 \times 2 \times 9}}{729-324}$
$\begin{aligned}
&=\frac{4 \times 3 \times 3+4 \times 3 \sqrt{6}+4 \times 3 \sqrt{6}+4 \times 2 \times 3}{405} \\
&=\frac{36+12 \sqrt{6}+12 \sqrt{6}+24}{405}=\frac{60+24 \sqrt{6}}{405}
\end{aligned}$
(ii)$\frac{5 \sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{(5 \sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})} \times \frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})}$
$=\frac{5 \sqrt{3} \times \sqrt{3}+\sqrt{2} \times \sqrt{3}-5 \sqrt{3} \times \sqrt{2}-\sqrt{2}^{2}}{\sqrt{3}^{2}-\sqrt{2}^{2}}$
$=\frac{5 \times 3+\sqrt{6}-5 \sqrt{6}-2}{3-2}=\frac{13-4 \sqrt{6}}{1}=13-4 \sqrt{6}$
(iii)$\begin{aligned}
\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}} &=\frac{(2 \sqrt{6}-\sqrt{5}) \times 3 \sqrt{5}+2 \sqrt{6}}{(3 \sqrt{5}-2 \sqrt{6}) \times 3 \sqrt{5}+2 \sqrt{6}} \\
&=\frac{2 \sqrt{6} \times 3 \sqrt{5}-3 \sqrt{5} \sqrt{5}-(2 \sqrt{6})^{2}-2 \sqrt{5} \times \sqrt{6}}{(3 \sqrt{5})^{2}-(2 \sqrt{6})^{2}} \\
&=\frac{6 \sqrt{30}-3 \times 5-4 \times 6-2 \sqrt{30}}{9 \times 5-4 \times 6}=\frac{4 \sqrt{30}-39}{45-24}=\frac{4 \sqrt{30}-39}{21}
\end{aligned}$
(iv)$\begin{aligned}
\frac{\sqrt{5}}{\sqrt{6}+2}-\frac{\sqrt{5}}{\sqrt{6}-2} &=\frac{\sqrt{5}(\sqrt{6}-2)-\sqrt{5}(\sqrt{6}+2)}{(\sqrt{6}+2)(\sqrt{6}-2)} \\
&=\frac{\sqrt{30}-2 \sqrt{5}-\sqrt{30}-2 \sqrt{5}}{\sqrt{6}^{2}-2^{2}}=\frac{-4 \sqrt{5}}{6-4}=\frac{-4 \sqrt{5}}{2}=-2 \sqrt{5}
\end{aligned}$

Question $3 .$
Find the value of a and $\mathrm{b}$ if $\frac{\sqrt{7}-2}{\sqrt{7+2}}=a \sqrt{7}+b$.
Solution:
$\begin{aligned}
\frac{\sqrt{7}-2}{\sqrt{7}+2}=a \sqrt{7}+b \\
\text {. L.H.S } &=\frac{\sqrt{7}-2 \times \sqrt{7}-2}{\sqrt{7}+2 \times \sqrt{7}-2}=\frac{(\sqrt{7}-2)^{2}}{\sqrt{7}^{2}-2^{2}}=\frac{\sqrt{7}^{2}-2 \sqrt{7} \times 2+2^{2}}{7-4} \\
\frac{-4 \sqrt{7}}{3}+\frac{11}{3} &=\frac{7-4 \sqrt{7}+4}{3}=\frac{11-4 \sqrt{7}}{3}=\frac{11}{3}-\frac{4 \sqrt{7}}{3} \\
\therefore a \sqrt{7} &=\frac{-4 \sqrt{7}}{3} \Rightarrow a=\frac{-4}{3}
\end{aligned}$

$b=\frac{11}{3}$

Question $4 .$
If $x=\sqrt{5}+2$, then find the value of $x^{2}+\frac{1}{x^{2}}$.
Solution:
If $x=\sqrt{5}+2$
$\begin{gathered}
\frac{1}{x}=\frac{1}{\sqrt{5}+2}=\frac{1}{\sqrt{5}+2} \times \frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{\sqrt{5}{ }^{2}-2^{2}}=\frac{\sqrt{5}-2}{5-4}=\frac{\sqrt{5}-2}{1} \\
x+\frac{1}{x}=\sqrt{5}+2+\sqrt{5}-22=2 \sqrt{5} \Rightarrow\left(x+\frac{1}{x}\right)^{2}=(2 \sqrt{5})^{2}=4 \times 5=20
\end{gathered}$

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:
$\begin{aligned}
\frac{8-5 \sqrt{2}}{3-2 \sqrt{2}} &=\frac{(8-5 \sqrt{2}) \times(3+2 \sqrt{2})}{(3-2 \sqrt{2}) \times(3+2 \sqrt{2})}=\frac{24-15 \sqrt{2}+16 \sqrt{2}-10 \times 2}{3^{2}-(2 \sqrt{2})^{2}} \\
&=\frac{24+\sqrt{2}-20}{9-4 \times 2}=\frac{4+\sqrt{2}}{1}=4+1.414=5.414
\end{aligned}$