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

Answer:

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

Answer:

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}

Answer:

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}

Answer:

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

Answer:

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

Answer:

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}

 

 

 

 

 

 

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