First half of Exercise 3(a)…

Question 12: Find the rationalising factors of the following:

(i) $\sqrt{108}$

$\sqrt{108} = \sqrt{2^2 \times 3^3} = 6 \sqrt{3}$.

We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt{3} =$ $3^{\frac{1}{2}}$ the rationalizing factor should be $3^{(1-\frac{1}{2})} = 3^{\frac{1}{2}} = \sqrt{3}$

(ii) $\sqrt[5]{486}$

$\sqrt[5]{486} = \sqrt[5]{2 \times 3^5} = 3 \sqrt[5]{2}$.

We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt[5]{2} =$ $2^{\frac{1}{5}}$ the rationalizing factor should be $2^{(1-\frac{1}{5})} = 2^{\frac{4}{5}} = \sqrt[5]{16}$

(iii) $\sqrt{2}+\sqrt{3}+\sqrt{5}$

We find that

$( \sqrt{2}+\sqrt{3}+\sqrt{5}) ( \sqrt{2}+\sqrt{3}-\sqrt{5}) = ( \sqrt{2}+\sqrt{3})^2 - (\sqrt{5})^2 = 2 + 3 + 2 \sqrt{6} - 5 = 2 \sqrt{6}$

Rationalizing factor of $\sqrt{6}$ is $\sqrt{6}$.

Hence $\sqrt{6}( \sqrt{2}+\sqrt{3}-\sqrt{5})$ is the rationalizing factor of  $\sqrt{2}+\sqrt{3}+\sqrt{5}$

(iv) $\sqrt{3}+ \sqrt{8+2\sqrt{15}}$

We have $\sqrt{3}+ \sqrt{8+2\sqrt{15}} = \sqrt{3}+ \sqrt{5 + 3+2\sqrt{5 \times 3}} = \sqrt{3}+ \sqrt{(\sqrt{5} + \sqrt{3})^2}$

$= \sqrt{3} +\sqrt{5}+\sqrt{3} = 2 \sqrt{3}+ \sqrt{5}$

The conjugate of $(2 \sqrt{3}+ \sqrt{5})$ is $(2 \sqrt{3}- \sqrt{5})$

Therefore the rationalizing factor of $\sqrt{3}+ \sqrt{8+2\sqrt{15}}$ is $(2\sqrt{3}-\sqrt{5})$.

(v) $\sqrt[3]{5}$

We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt[3]{5} =$ $5^{\frac{1}{3}}$ the rationalizing factor should be $5^{(1-\frac{1}{3})} = 5^{\frac{2}{3}} = \sqrt[3]{25}$

Question 13: Rationalize the denominator and simplify

(i)    $\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$

$=$ $\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$ $\times$ $\frac{6-4\sqrt{2}}{6-4\sqrt{2}}$

$=$ $\frac{36 + 32 -48\sqrt{2}}{36-32}$

$=$ $\frac{78-48\sqrt{2}}{4}$

$= 17 - 12\sqrt{2}$

(ii) $\frac{b^2}{\sqrt{a^2+b^2} +a}$

$=$ $\frac{b^2}{\sqrt{a^2+b^2} +a}$ $\times$ $\frac{\sqrt{a^2+b^2} -a}{\sqrt{a^2+b^2} -a}$

$=$ $\frac{b^2(\sqrt{a^2+b^2)}}{a^2+b^2-a^2}$

$=$ $\sqrt{a^2+b^2} - a$

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

$=$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+ \sqrt{2}}$ $\times$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}- \sqrt{2}}$

$=$ $\frac{3+2-2\sqrt{6}}{3-2}$

$=$ $5 - 2\sqrt{6}$

(iv) $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$

$=$ $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$ $\times$ $\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}$

$=$ $\frac{6\sqrt{30} - 15 + 24 - 2 \sqrt{30}}{45-24}$

$=$ $\frac{4\sqrt{3}+9}{21}$

(v)  $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$

$=$ $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$ $\times$ $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}}$

$=$ $\frac{a+b+\sqrt{a^2-b^2} + \sqrt{a^2-b^2}+a-b}{a+b-a+b}$

$=$ $\frac{2(a+\sqrt{a^2-b^2})}{2b}$

$=$ $\frac{a+\sqrt{a^2-b^2}}{b}$

Question 14: Simplify

(i) $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$ $+$ $\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$

$=$ $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$ $\times$ $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$ $+$ $\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$ $\times$ $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=$ $\frac{18-6\sqrt{6} - 6\sqrt{6}+12}{18-12}$ $+$ $\frac{\sqrt{12}(\sqrt{3}+\sqrt{2})}{1}$

$=$ $\frac{30-12\sqrt{6}}{6}$ $+$ $\sqrt{12} (\sqrt{3}+\sqrt{2})$

$=$ $5 - 2\sqrt{6} + 6 + 2\sqrt{6}$

$=$ $11$

(ii) $\frac{1}{2+\sqrt{3}}$ $+$ $\frac{2}{\sqrt{5} - \sqrt{3}}$ $+$ $\frac{1}{2-\sqrt{5}}$

$=$ $\frac{1}{2+\sqrt{3}}$ $\times$ $\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $+$ $\frac{2}{\sqrt{5} - \sqrt{3}}$ $\times$ $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ $+$ $\frac{1}{2-\sqrt{5}}$ $\times$ $\frac{2+\sqrt{5}}{2+\sqrt{5}}$

$=$ $\frac{2-\sqrt{3}}{4-3}$ $+$ $\frac{2(\sqrt{5}+\sqrt{3})}{5-3}$ $+$ $\frac{2+\sqrt{5}}{-1}$

$= 2 -\sqrt{3}+\sqrt{5}+\sqrt{3}-2-\sqrt{5}$

$= 0$

(iii) $\frac{2}{\sqrt{5}+\sqrt{3}}$ $+$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $-$ $\frac{3} {\sqrt{5}+\sqrt{2}}$

$=$ $\frac{2}{\sqrt{5}+\sqrt{3}}$ $\times$ $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ $+$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $\times$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $+$ $\frac{3} {\sqrt{5}+\sqrt{2}}$ $\times$ $\frac{\sqrt{5}-\sqrt{2}} {\sqrt{5}-\sqrt{2}}$

$=$ $\frac{2(\sqrt{5}-\sqrt{3})}{2}$ $+$ $\frac{\sqrt{3}-\sqrt{2}}{1}$ $-$ $\frac{3(\sqrt{5}-\sqrt{2})}{3}$

$=$ $\sqrt{5} - \sqrt{3} + \sqrt{3} - \sqrt{2} - \sqrt{5} + \sqrt{2}$

$= 0$

(iv) $\frac{1}{2-\sqrt{3}}$ $-$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $+$ $\frac{5}{3-\sqrt{2}}$

$=$ $\frac{1}{2-\sqrt{3}}$ $\times$ $\frac{2+\sqrt{3}}{2+\sqrt{3}}$ $-$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $\times$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $+$ $\frac{5}{3-\sqrt{2}}$ $\times$ $\frac{3+\sqrt{2}}{3+\sqrt{2}}$

$=$ $\frac{2+\sqrt{3}}{4-3}$ $-$ $\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{5(\sqrt{3}+\sqrt{2})}{9-2}$

$=$ $2 + \sqrt{3}-\sqrt{3}+\sqrt{2}$ $+$ $\frac{15+5\sqrt{2}}{7}$

$=$ $\frac{14+7\sqrt{2}+15+5\sqrt{2}}{7}$

$=$ $\frac{29+12\sqrt{2}}{7}$

(v)  $\frac{4\sqrt{3}}{2+\sqrt{3}}$ $-$ $\frac{30}{4\sqrt{3}-3\sqrt{2}}$ $-$ $\frac{3\sqrt{2}}{3+2\sqrt{2}}$

$=$ $\frac{4\sqrt{3}}{2+\sqrt{3}}$ $\times$ $\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $-$ $\frac{30}{4\sqrt{3}-3\sqrt{2}}$ $\times$ $\frac{4\sqrt{3}+3\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$ $-$ $\frac{3\sqrt{2}}{3+2\sqrt{2}}$ $\times$ $\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$=8\sqrt{3} - 12 - 4\sqrt{3}-3\sqrt{2}-9\sqrt{2}+12$

$= 4\sqrt{3}-12\sqrt{2}$

$= 4(\sqrt{3}-3\sqrt{2})$

Question 15: Determine rational numbers $a$ and $b$

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

$\Rightarrow$ $\frac{3+\sqrt{2}}{3-\sqrt{2}}$ $\times$ $\frac{3+\sqrt{2}}{3+\sqrt{2}}$ $=$ $a + b\sqrt{2}$

$\Rightarrow$ $\frac{9+2+6\sqrt{2}}{9-2}$ $=$ $a + b\sqrt{2}$

$\Rightarrow$ $\frac{11}{7} + \frac{6}{11}$ $\sqrt{2}$ $=$ $a + b\sqrt{2}$

$\Rightarrow$ $a =$ $\frac{11}{7}$ and $b =$ $\frac{6}{11}$

(ii) $\frac{5+3\sqrt{3}}{7+4\sqrt{3}}$ $=$ $a + b\sqrt{3}$

$\Rightarrow$ $\frac{5+3\sqrt{2}}{7+3\sqrt{3}}$ $\times$ $\frac{7-4\sqrt{3}}{7-4\sqrt{3}}$ $=$ $a + b\sqrt{3}$

$\Rightarrow$ $\frac{35+21\sqrt{3}-20\sqrt{3}-36}{49-48}$ $=$ $a + b\sqrt{3}$

$\Rightarrow$ $-1+\sqrt{3}$ $=$ $a + b\sqrt{3}$

$\Rightarrow$ $a = -1$ and $b = 1$

(iii) $\frac{1+\sqrt{48}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{72} - \sqrt{108}+\sqrt{8}+2}$ $=$ $a + b\sqrt{3}$

Let’s first simplify

$\frac{1+\sqrt{48}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{72} - \sqrt{108}+\sqrt{8}+2}$

$=$ $\frac{1+\sqrt{2^4 \times 3}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{2^3 \times 3^2} - \sqrt{2^2 \times 3^3}+\sqrt{2^3}+2}$

$=$ $\frac{1+3\sqrt{3}}{5\sqrt{3} + 4\sqrt{2} - 6\sqrt{2} - 6\sqrt{3}+2\sqrt{2}+2}$

$=$ $\frac{1+4\sqrt{3}}{2-\sqrt{3}}$

$=$ $\frac{1+4\sqrt{3}}{2-\sqrt{3}}$ $\times$ $\frac{2+\sqrt{3}}{2+\sqrt{3}}$

$=$ $\frac{2+8\sqrt{3}+\sqrt{3}+12}{4-3}$

$= 14+9\sqrt{3}$

Now comparing,  $14+9\sqrt{3} = a + b\sqrt{3}$

$\Rightarrow a = 14 \ and \ b = 9$

Question 16: If $x$ $=$ $\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $y$ $=$ $\frac{\sqrt{2}-1}{\sqrt{2}+1}$, find $x^2+xy+y^2$

$x^2+xy+y^2$

$=$ $(\frac{\sqrt{2}+1}{\sqrt{2}-1})^2$ $+$ $(\frac{\sqrt{2}+1}{\sqrt{2}-1})(\frac{\sqrt{2}-1}{\sqrt{2}+1})$ $+$ $(\frac{\sqrt{2}-1}{\sqrt{2}+1})^2$

$=$ $\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$ $+ 1 +$ $\frac{2+1-2\sqrt{2}}{2+1+2\sqrt{2}}$

$=$ $\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$ $+ 1 +$ $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$

$=$ $\frac{9+9+12\sqrt{2} + 9 + 8 - 12\sqrt{2}}{9-8}$ $+1$

$= 34+1 = 35$

Question 17: If $x$ $=$ $\frac{1}{3-2\sqrt{2}}$ and $y$ $=$ $\frac{1}{3+2\sqrt{2}}$, find $xy^2+x^2y$

$xy^2+x^2y$

$= ($ $\frac{1}{3-2\sqrt{2}})(\frac{1}{3+2\sqrt{2}})^2$ $+ ($ $\frac{1}{3-2\sqrt{2}})^2(\frac{1}{3+2\sqrt{2}})$

$=$ $\frac{1}{9-8}(\frac{1}{3+2\sqrt{2}})$ $+$ $\frac{1}{9-8}(\frac{1}{3-2\sqrt{2}})$

$=$ $\frac{1}{3+2\sqrt{2}}$ $+$ $\frac{1}{3-2\sqrt{2}}$

$=$ $\frac{3-2\sqrt{2}+3+2\sqrt{2}}{9-8}$

$= 6$

Question 18: If $x = 2 + \sqrt{3}$, find the value of $x^3 +$ $\frac{1}{x^3}$

$x = 2 + \sqrt{3}$

Therefore $\frac{1}{x}$ $=$ $\frac{1}{2 + \sqrt{3}}$ $\times$ $\frac{2 - \sqrt{3}}{2 - \sqrt{3}}$ $=$ $2-\sqrt{3}$

Hence $x +$ $\frac{1}{x}$ $= 2 + \sqrt{3} + 2-\sqrt{3} = 4$

$x^3 +$ $\frac{1}{x^3}$ $=$ $($ $x+$ $\frac{1}{x})^3$ $- 3(x+$ $\frac{1}{x}$ $) = 4^3 - 3 \times 4 = 52$

Question 19: If $x = 3 + \sqrt{8}$, find the value of $x^2 +$ $\frac{1}{x^2}$

$x = 3 + \sqrt{8}$

Therefore $\frac{1}{x}$ $=$ $\frac{1}{3 + \sqrt{8}}$ $\times$ $\frac{3 - \sqrt{8}}{3 - \sqrt{8}}$ $=$ $3-\sqrt{8}$

Hence $x +$ $\frac{1}{x}$ $= 3 + \sqrt{8} + 3-\sqrt{8} = 6$

$x^2 +$ $\frac{1}{x^2}$ $=$ $($ $x+$ $\frac{1}{x})^2$ $- 2$ $= 6^2 - 2 = 34$

Question 20: If $x =$ $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $y =$ $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ , find the value of $3x^2+4xy-3y^2$

$3x^2+4xy-3y^2$

$= 3($ $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}}$ $)^2 + 4 ($ $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}}$ $)($ $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ $) - 3 ($ $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ $)^2$

$= 3 ($ $\frac{7+\sqrt{2}}{7-\sqrt{2}}$ $) + 4 - 3 ($ $\frac{7-\sqrt{2}}{7+\sqrt{2}}$ $)$

$= 3 ($ $\frac{49+40+28\sqrt{10}-49-40+28\sqrt{10}}{49-40}$ $)$

$=$ $\frac{56\sqrt{10}+12}{3}$

Question 21: If $x =$ $\frac{\sqrt{3}+1}{2}$, find the value of $4x^3+2x^2-8x+7$

$4x^3+2x^2-8x+7$

$= 4 \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^3 + 2 \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^2 - 8\Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big] + 7$

$= \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^2 \Big[ ($ $\frac{4\sqrt{3}+4}{2}$ $) + 2 \Big] - 8\Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big] + 7$

$=$ $(\frac{3+1+2\sqrt{3}}{2})(\frac{4\sqrt{3}+4+4}{2})$ $- 8$ $(\frac{\sqrt{3}+1}{2})$ $+ 7$

$= (4+2\sqrt{3})$ $(\frac{\sqrt{3}+2}{2})$ $- 8$ $(\frac{\sqrt{3}+1}{2})$ $+ 7$

$=$ $\frac{4\sqrt{3}+6+8+4\sqrt{3}-8\sqrt{3}-8+14}{2}$

$= 10$

Question 22: Given $\sqrt{3}= 1.732, \sqrt{5}= 2.236, \sqrt{2}=1.4142, \sqrt{6}=2.4495$ and $\sqrt{10}=3.162$, find

(i)   $\frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}$ $+$ $\frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}$

$=$ $\frac{\sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} + \sqrt{5} + \sqrt{3} -\sqrt{10} -\sqrt{6}}{5-3}$

$=$ $\frac{2\sqrt{5}-2\sqrt{6}}{2}$

$= \sqrt{5}-\sqrt{6}$

$= 2.236 - 2.4495 = -0.213$

(ii) $\frac{6}{\sqrt{5}-\sqrt{3}}$

$=$ $\frac{6}{\sqrt{5}-\sqrt{3}}$ $\times$ $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$=$ $\frac{6(\sqrt{5}+\sqrt{3})}{5-3}$

$= 3 (\sqrt{5}+\sqrt{3})$

$= 3(2.236+1.732) = 11.904$

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

$=$ $\frac{2+\sqrt{3}}{2-\sqrt{3}}$ $\times$ $\frac{2+\sqrt{3}}{2+\sqrt{3}}$ $+$ $\frac{2-\sqrt{3}}{2+\sqrt{3}}$ $\times$ $\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $\times$ $\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $\times$ $\frac{\sqrt{3}-1}{\sqrt{3}-1}$

$= 4+3+4\sqrt{3}+4+3-4\sqrt{3} +$ $\frac{3+1-2\sqrt{3}}{2}$

$= 14 + (2-\sqrt{3})$

$= 14 + (2-1.732) = 14.268$

Question 23: Rationalize and simplify:

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

$=$ $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ $\times$ $\frac{1+\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}$

$=$ $\frac{1+\sqrt{2}+\sqrt{3}}{1+2+2\sqrt{2}-3}$

$=$ $\frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$ $\times$ $\frac{\sqrt{2}}{\sqrt{2}}$

$=$ $\frac{\sqrt{2}+2+\sqrt{6}}{4}$

(ii) $\frac{1}{1+\sqrt{5}+\sqrt{3}}$

$=$ $\frac{1}{1+\sqrt{5}+\sqrt{3}}$ $\times$ $\frac{1+\sqrt{5} - \sqrt{3}}{1+\sqrt{5} -\sqrt{3}}$

$=$ $\frac{1+\sqrt{5}-\sqrt{3}}{1+5+2\sqrt{5}-3}$

$=$ $\frac{1+\sqrt{5}-\sqrt{3}}{3+2\sqrt{5}}$ $\times$ $\frac{3-2\sqrt{5}}{3-2\sqrt{5}}$

$=$ $\frac{3+3\sqrt{5}-3\sqrt{3}-2\sqrt{5}-10+2\sqrt{15} }{9-20}$

$=$ $\frac{7-\sqrt{5}+3\sqrt{3}-2\sqrt{15}}{11}$

(iii) $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$

$=$ $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$ $\times$ $\frac{{\sqrt{2}+\sqrt{3}+\sqrt{5}}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

$=$ $\frac{2+\sqrt{6}+\sqrt{10}}{2+3+2\sqrt{6}-5}$

$=$ $\frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}}$

$=$ $\frac{2\sqrt{6}+6+\sqrt{60}}{12}$

$=$ $\frac{\sqrt{6}+3+\sqrt{15}}{6}$