Exercise 2.6 - Chapter 2 - Real Numbers - 9th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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

$\begin{aligned}
& =3 \sqrt{5 \times 5 \times 3}+5 \sqrt{3 \times 2 \times 2 \times 2 \times 2}-\sqrt{3 \times 3 \times 3 \times 3 \times 3} \\
& =3 \times 5 \sqrt{3}+5 \times 2 \times 2 \sqrt{3}-3 \times 3 \sqrt{3} \\
& =15 \sqrt{3}+20 \sqrt{3}-9 \sqrt{3} \\
& =(15+20-9) \sqrt{3}=26 \sqrt{3}
\end{aligned}$

(iv) $5 \sqrt[3]{40}+2 \sqrt[3]{625}-3 \sqrt[3]{320}$
$
\begin{aligned}
& =5 \sqrt[3]{2^3 \times 5}+2 \sqrt[3]{5^3 \times 5}-3 \sqrt[3]{2^3 \times 2^3 \times 5} \\
& =5 \times 2 \times \sqrt[3]{5}+2 \times 5 \sqrt[3]{5}-3 \times 2 \times 2 \sqrt[3]{5} \\
& =10 \sqrt[3]{5}+10 \sqrt[3]{5}-12 \sqrt[3]{5} \\
& =(10+10-12) \sqrt[3]{5}=8 \sqrt[3]{5} \\
&
\end{aligned}
$

Question 2.
Simplify the following using multiplication and division properties of surds :
(i) $\sqrt{3} \times \sqrt{5} \times \sqrt{2}$
(ii) $\sqrt{35} \div \sqrt{7}$
(iii) $\sqrt[3]{27} \times \sqrt[3]{8} \times \sqrt[3]{125}$
(iv) $(7 \sqrt{a}-5 \sqrt{b})(7 \sqrt{a}+5 \sqrt{b})$
(v) $\left[\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}\right] \div \sqrt{\frac{16}{81}}$

Solution:
$\sqrt{3} \times \sqrt{5} \times \sqrt{2}=\sqrt{3 \times 5 \times 2}=\sqrt{30}$
$\sqrt{35} \div \sqrt{7}=\sqrt{\frac{35}{7}}=\sqrt{5}$
(iii) $\sqrt[3]{27} \times \sqrt[3]{8} \times \sqrt[3]{125}=\sqrt[3]{27 \times 8 \times 125}=\sqrt[3]{3^{3} \times 2^{3} \times 5^{3}}=3 \times 2 \times 5=30$
(iv) $(7 \sqrt{a}-5 \sqrt{b})(7 \sqrt{a}+5 \sqrt{b})=(7 \sqrt{a})^{2}-(5 \sqrt{b})^{2}=49 a-25 b$
(v) $\begin{aligned}
{\left[\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}\right]+\sqrt{\frac{16}{81}} } & {\left[\sqrt{\frac{15^{2}}{27^{2}}}-\sqrt{\frac{5^{2}}{12^{2}}}\right] \times \sqrt{\frac{9^{2}}{4^{2}}} } \\
&=\left(\frac{15}{27}-\frac{5}{12}\right) \times \frac{9}{4}=\left(\frac{5}{9}-\frac{5}{12}\right) \times \frac{9}{4} \\
&=\left(\frac{20-15}{36}\right) \times \frac{9}{4}=\frac{5}{36} \times \frac{9}{4}=\frac{5}{6}
\end{aligned}$

Question $3 .$
If $\sqrt{7}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236, \sqrt{10}=3.162$, 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)$\begin{aligned}
\sqrt{40}-\sqrt{20} &=\sqrt{4 \times 10}-\sqrt{2 \times 10}=\sqrt{(4-2) 10}=\sqrt{2 \times 10}=\sqrt{2} \times \sqrt{10} \\
&=1.414 \times 3.162=4.471068=4.471
\end{aligned}$
(ii)$\begin{aligned}
\sqrt{300}+\sqrt{90}-\sqrt{8} &=\sqrt{3 \times 100}+\sqrt{9 \times 10}+\sqrt{4 \times 2}=10 \sqrt{3}+3 \sqrt{10}+2 \sqrt{2} \\
&=10 \times 1.732+3 \times 3.162+2 \times 1.414 \\
&=17.32+9.486+2.828=29.634
\end{aligned}$

Question $4 .$
Arrange surds in descending order :
(i) $\sqrt[3]{5}, \sqrt[8]{4}, \sqrt[6]{3}$
(ii) $\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$
Solution:
(i) $\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$
$5^{\frac{1}{3}}$
$\therefore$ The order of the surds $\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$ are $3,9,6$.
$4^{\frac{1}{9}}$
$3^{\frac{1}{6}}$ l.c.m of $3,9,6$ is 18
$\therefore \frac{1}{3}=\frac{1 \times 6}{3 \times 6}=\frac{6}{18}$
$\frac{1}{9}=\frac{1 \times 2}{9 \times 2}=\frac{2}{18}$
$\frac{1}{6}=\frac{1 \times 3}{6 \times 3}=\frac{3}{18}$
$\left(5^{\frac{1}{3}}\right)=5^{\frac{6}{18}}=\left(5^{6}\right)^{\frac{1}{18}}=(15625)^{\frac{1}{18}}$

$\begin{aligned}
\left(4^{\frac{1}{9}}\right) &=4^{\frac{2}{18}}=\left(4^{2}\right)^{\frac{1}{18}}=16^{\frac{1}{18}} \\
\left(3^{\frac{1}{6}}\right) &=3^{\frac{3}{18}}=\left(3^{3}\right)^{\frac{1}{18}}=27^{\frac{1}{18}}
\end{aligned}$
$\therefore$ The descending order of $\sqrt[8]{5}, \sqrt[9]{4}, \sqrt[6]{3}$ is $(15625)^{\frac{1}{18}}>(27)^{\frac{1}{18}}>16^{\frac{1}{16}}$ i.e. $\sqrt[3]{5}>\sqrt[6]{3}>\sqrt[9]{4}$
(ii) $\sqrt[2]{\sqrt[3]{5}}, \sqrt[5]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$
The order of the surds $\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$ are $6,12,4$
l.c.m of $6,12,4$ is 12
$\begin{aligned}
&\sqrt[2]{\sqrt[3]{5}}=5^{\frac{1}{6}}=5^{\frac{1 \times 2}{6 \times 2}}=5^{\frac{2}{12}}=\left(5^{2}\right)^{\frac{1}{12}}=25^{\frac{1}{12}} \\
&\sqrt[3]{\sqrt[4]{7}}=7^{\frac{1}{12}} \\
&\sqrt{\sqrt{3}}=3^{\frac{1}{4}}=3^{\frac{1 \times 3}{4 \times 3}}=3^{\frac{3}{12}}=\left(3^{3}\right)^{\frac{1}{12}}=27^{\frac{1}{12}}
\end{aligned}$
$\therefore$ The ascending order of the surds
$\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$ is $7^{\frac{1}{12}}<25^{\frac{1}{2}}<27^{\frac{1}{2}}$, that is $\sqrt[3]{\sqrt[4]{7}}<\sqrt[2]{\sqrt[3]{5}}<\sqrt{\sqrt{3}}$
 

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 $4 \sqrt[3]{21}+(-3 \sqrt[3]{21})=\sqrt[3]{21}$
(ii) $\mathrm{Yes}$
(iii) Yes
(iv) Yes
$\sqrt[7]{25}-6 \sqrt[4]{25}=\sqrt[4]{25}(7-6)=\sqrt[4]{25}$
$\sqrt[3]{5} \times \sqrt[3]{4}=\sqrt[3]{20}$, a pure surd
$=\frac{\sqrt{2 \times 5}}{\sqrt{2}}=\frac{\sqrt{2} \times \sqrt{5}}{\sqrt{2}}=\sqrt{5}$

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 $(5-\sqrt{3})+(5+\sqrt{3})=10$, a rational number
(ii) Yes $(5+\sqrt[3]{7})-(-6+\sqrt[3]{7})=11$, a rational number
(iii) Yes $(5+\sqrt{3})(5-\sqrt{3})=25-3=22$, a rational number
(iv) Yes $\frac{5 \sqrt{3}}{\sqrt{3}}=5$, a rational number