Exercise 1(C)
Question 1
(i) $\sqrt{\frac{1}{3}}$
(ii) $\sqrt{\frac{5}{12}}$
(iii) $\sqrt{1 \frac{46}{75}}$
Sol :
(i) $\sqrt{\frac{1}{3}}$
Let be $\sqrt{\frac{1}{3}}$
$\Rightarrow \frac{1 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$ (Multiplying and dividing by $\sqrt{3}$)
$=\frac{\sqrt{3}}{3}$
(ii) $\sqrt{\frac{5}{12}}=\frac{\sqrt{5}}{\sqrt{2 \times 2 \times 3}}=\frac{\sqrt{5}}{2 \sqrt{3}}$
Multiplying and dividing by $\sqrt{3}$
$\Rightarrow \frac{\sqrt{5} \sqrt{3}}{2 \sqrt{3} \cdot \sqrt{3}} \Rightarrow \frac{\sqrt{5} \sqrt{3}}{2 \times 3}$
$\Rightarrow \frac{\sqrt{5} \sqrt{3}}{6}$
$\Rightarrow \frac{\sqrt{15}}{6}=\frac{1}{6} \sqrt{15}$
(iii) $\sqrt{1 \frac{46}{75}}$
$=\sqrt{\frac{1 \times 75+46}{75}}=\sqrt{\frac{121}{75}}$
$=\frac{\sqrt{121}}{\sqrt{75}}=\frac{\sqrt{11 \times 11}}{\sqrt{3 \times 5 \times 5}}$
$\Rightarrow \frac{11 \sqrt{3}}{5 \sqrt{3} \times \sqrt{3}}=\frac{11 \sqrt{3}}{5 \times 3}$
$=\frac{11}{15} \sqrt{3}$
Question 2
$\sqrt{112}-\sqrt{63}+\frac{224}{\sqrt{28}}$
Sol :
⇒$\sqrt{112}-\sqrt{63}+\frac{224}{\sqrt{28}}$
⇒$\sqrt{2 \times 2 \times 2 \times 2 \times 7}-\sqrt{3 \times 3+7}+\frac{224}{\sqrt{2 \times 2 \times 7}}$
⇒$ 2 \times 2 \sqrt{7}-3 \sqrt{7}+\frac{224}{2 \sqrt{7}}$
⇒$4 \sqrt{7}-3 \sqrt{7}+\frac{112}{\sqrt{7}}$
⇒$4 \sqrt{7}-3 \sqrt{7}+\frac{112 \sqrt{7}}{\sqrt{7} \times \sqrt{7}}$
⇒$4 \sqrt{7}-3 \sqrt{7}+\frac{112 \sqrt{7}}{7}$
⇒$4 \sqrt{7}-3 \sqrt{7}+16 \sqrt{7}$
⇒$(4-3+16) \sqrt{7}=17 \sqrt{7}$
Question 3
$\frac{4 \sqrt{18}}{\sqrt{12}}-\frac{8 \sqrt{75}}{\sqrt{32}}+\frac{9 \sqrt{2}}{3}$
Sol :
$\frac{4 \sqrt{18}}{\sqrt{12}}-\frac{8 \sqrt{75}}{\sqrt{32}}+\frac{9 \sqrt{2}}{3}$
$=\frac{4 \sqrt{2 \times 3 \times 3}}{\sqrt{2 \times 2 \times 3}}-\frac{8 \sqrt{3 \times 5 \times 5}}{\sqrt{2 \times 2 \times 2 \times 2 \times 2}}+\frac{9 \sqrt{2}}{\sqrt{3}}$
$=\frac{4 \times 3 \sqrt{2}}{2 \sqrt{3}}-\frac{8 \times 5 \sqrt{3}}{2 \times 2 \sqrt{2}}+\frac{9 \sqrt{2}}{\sqrt{3}}$
$=\frac{6 \sqrt{2}}{\sqrt{3}}-\frac{10 \sqrt{3}}{\sqrt{2}}+\frac{9 \sqrt{2}}{\sqrt{3}}$
$=\frac{6 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}-\frac{10 \sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}+\frac{9 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$
(Rationalising denominator of each)
$=\frac{6 \sqrt{6}}{3}-\frac{10 \sqrt{6}}{2}+\frac{9 \sqrt{6}}{3}$
$=2 \sqrt{6}-5 \sqrt{6}+3 \sqrt{6}$
$=5 \sqrt{6}-5 \sqrt{6}=0$
Question 4
(i) $\frac{1}{4-\sqrt{3}}$
(ii) $\frac{2}{\sqrt{5}+\sqrt{3}}$
(iii) $\frac{1}{2 \sqrt{5}-\sqrt{3}}$
(iv) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Sol :
(i) $\frac{1}{4-\sqrt{3}} \times \frac{4+\sqrt{3}}{4+\sqrt{3}}$
$=\frac{4+\sqrt{3}}{(4)^{2}-(\sqrt{3})^{2}}$
$=\frac{4+\sqrt{3}}{16-3}$
$=\frac{4+\sqrt{3}}{13}$
(ii) $\frac{2}{\sqrt{5}+\sqrt{3}}=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$
Rationalising factor of $(\sqrt{5}+\sqrt{3})$ is
$=\frac{2 (\sqrt{5}-\sqrt{3})}{5-3}=\frac{2 \sqrt{5}+2\sqrt{3}}{2}$
(iii) $\frac{1}{2 \sqrt{5}-\sqrt{3}}=\frac{1(2 \sqrt{5}+\sqrt{3})}{(2 \sqrt{5}-\sqrt{3})(2 \sqrt{5}+\sqrt{3})}$
Rationalising factor of $(2 \sqrt{5}-3) \text{ is }(2 \sqrt{5}+3)$
$=\frac{2 \sqrt{5}+\sqrt{3}}{(2 \sqrt{5})^{2}-(\sqrt{3})^{2}}$
$=\frac{2 \sqrt{5}+\sqrt{3}}{20-3}$
$=\frac{2 \sqrt{5}+\sqrt{3}}{17}$
(iv) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
Rationalising factor of $(\sqrt{3-} \sqrt{2})$ is $(\sqrt{3}+\sqrt{2})$
$=\frac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$
$=\frac{3+2+2 \sqrt{3} \times \sqrt{2}}{3-2}$
$=\frac{5+2 \sqrt{6}}{1}$
$=5+2 \sqrt{6}$
Question 5
Rationalise the denominator and simplify :
(i) $\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$
(ii) $\frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}}$
Sol :
(i) $\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$
$=\frac{(4+\sqrt{5})(4+55)}{(4-\sqrt{5})(4+\sqrt{5})}+\frac{(4-\sqrt{5})(4-\sqrt{5})}{(4+\sqrt{5})(4-\sqrt{5})}$
Rationalising the denominator
$=\frac{(4+\sqrt{5})^{2}}{(4)^{2}-(\sqrt{5})^{2}}+\frac{(4-\sqrt{5})^{2}}{(4)^{2}-(\sqrt{5})^{2}}$
$=\frac{16+5+2 \times 4 \sqrt{5}}{16-5}+\frac{16+5-2 \times 4 \sqrt{5}}{16-5}$
$=\frac{21+8 \sqrt{5}}{11}+\frac{21-8 \sqrt{5}}{11}$
$=\frac{21+8 \sqrt{5}+21-8 \sqrt{5}}{11}$
$=\frac{42}{11}$
(ii) $\frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}}$
$=\frac{3(5+\sqrt{3})}{(5-\sqrt{3})(5+\sqrt{3})}+\frac{2(5-\sqrt{3})}{(5+\sqrt{3})(5-\sqrt{3})}$
Rationalising the denominator
$=\frac{15+3 \sqrt{3}}{25-3}+\frac{10-2 \sqrt{3}}{25-3}$
$=\frac{15+3 \sqrt{3}}{22}+\frac{10-2 \sqrt{3}}{22}$
$=\frac{15+3 \sqrt{3}+10-2 \sqrt{3}}{22}$
$=\frac{25+\sqrt{3}}{22}$
Question 6
Find the values of a and b if
(i) $\frac{3+\sqrt{2}}{3-\sqrt{2}}=a=a+b\sqrt{2}$
(ii) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}$
(iii) $\frac{\sqrt{7}-1}{\sqrt{7+1}}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b \sqrt{7}$
Sol :
(i) $\frac{3+\sqrt{2}}{3-\sqrt{2}}=a=a+b\sqrt{2}$
$=\frac{(3+\sqrt{2})(3+ \sqrt{2})}{(3-\sqrt{2})(3+\sqrt{3})}=a+b \sqrt{2}$
$=\frac{(3+\sqrt{2})^{2}}{9-2}=a+b \sqrt{2}$
$=\frac{9+2+6 \sqrt{2}}{7}=a+b \sqrt{9}$
$=\frac{11+6 \sqrt{2}}{7}=a+b \sqrt{2}$
$=\frac{11}{7}+\frac{6}{7} \sqrt{2}=a+b\sqrt{2}$
comparing
$a=\frac{11}{7}, b=\frac{6}{7}$
(ii) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}$
$=\frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{(7+4 \sqrt{3})(7-4 \sqrt{3})}=a+b \sqrt{3}$
(Rationalising the denominator)
$=\frac{35-20 \sqrt{3}+14 \sqrt{3}-8 \times 3}{(7)^{2}-(4 \sqrt{3})^{2}}=a+b \sqrt{3}$
$=\frac{35-24-6 \sqrt{3}}{49-48}=a+b \sqrt{3}$
$=\frac{11-6 \sqrt{3}}{1}=a+b \sqrt{3}$
$=11-6 \sqrt{3}=a+b \sqrt{3}$
we get a=11, b=-6
(iii) $\frac{\sqrt{7}-1}{\sqrt{7+1}}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b \sqrt{7}$
$=\frac{(\sqrt{7}-1)^2-(\sqrt{7}+1)^2}{(\sqrt{7}+1)(\sqrt{7}-1)}=a+b\sqrt{7}$
$\frac{(7+1-2\sqrt{7})-(7+1+2\sqrt{7})}{7-1}=a+b\sqrt{7}$
$=\frac{8-2 \sqrt{7}-8-2 \sqrt{2}}{6}=a+b \sqrt{7}$
$=\frac{-4 \sqrt{7}}{6}=a+b\sqrt{7}$
$\frac{-2}{3} \sqrt{7}=a+b \sqrt{7}$
we get
a=0 , $b=\frac{-2}{3}$
Question 7
Rationalise the denominator of $\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$
Sol :
$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{5}}$
Rationalising denominator
$=\frac{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}{[(\sqrt{2}+\sqrt{3})+\sqrt{5})(\sqrt{2}+\sqrt{3})-\sqrt{5}]}$
$=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{5})^{2}}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2 \sqrt{6}-5}$
$=\frac{\sqrt{2}+\sqrt{3}-\sqrt5}{2 \sqrt{6}}$
Rationalising
$=\frac{(\sqrt{2}+\sqrt{3}-\sqrt{5}) \sqrt{6}}{2 \times \sqrt{6} \times \sqrt{6}}$
$=\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{2 \times 6}$
$=\frac{\sqrt{4 \times 3}+\sqrt{9 \times 2}-\sqrt{30}}{12}$
$=\frac{2 \sqrt{3}+3 \sqrt{2}-\sqrt{30}}{12}$
Question 8
Taking √2=1.414 and √3=1.732 , find without using tables or long division, the value of
(a) $\frac{1}{3-\sqrt{2}}$
(b) $\frac{2}{\sqrt{3-\sqrt{2}}}$
Sol :
(i) $\sqrt{2}=1.414 , \sqrt{3}=1.732$
$\frac{1}{3-\sqrt{2}}=\frac{1(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
Rationalising the denominator
$=\frac{3+\sqrt{2}}{(3)^{2}-(52)^{2}}$
$=\frac{3+\sqrt{2}}{(3)^{2}(\sqrt{2})^{2}}$
$=\frac{3+\sqrt2}{9-2}$
$=\frac{3+\sqrt{2}}{7}$
$=\frac{3+1.414}{7}$
$=\frac{4.414}{7}$
=0.631
(ii) $\frac{2}{\sqrt{3-\sqrt{2}}}=\frac{2(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
Rationalising denominator
$=\frac{2(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$
$=\frac{2(\sqrt{3}+\sqrt{2})}{3-2}$
$=2(\sqrt{3}+\sqrt{2})$
=2(1.732+1.414)
$=2 \times 3.146$
=6.292
Question 9
Express $\frac{3-5\sqrt{5}}{3+\sqrt{5}}$ in the form $(a\sqrt{5}– b)$ where a and b are simple fractions.
Sol :
$\frac{3-5\sqrt{5}}{3+\sqrt{5}}=\frac{(3-5\sqrt{5})(3-2 \sqrt{5})}{(3+2\sqrt{5})(3-3\sqrt{5})}$
$=\frac{9-6\sqrt5-15 \sqrt{5}+10 \times 5}{(3)^{2}-(2\sqrt5)^{2}}$
$=\frac{9-21 \sqrt{5}+20}{9-20}$
$=\frac{59-21\sqrt5}{-11}$
$=\frac{-59}{11}+\frac{21}{11} \sqrt{5}$
=It is in the form of $a\sqrt{5}-b$
Question 10
Prove that $\frac{1}{\sqrt{2}-1}+\frac{2}{\sqrt{3}+1}=\sqrt{2}+\sqrt{3}$
Sol :
$\frac{1}{\sqrt{2}-1}+\frac{2}{\sqrt{3}+1}=\sqrt{2}+\sqrt{3}$
L.H.S$=\frac{1}{\sqrt{2-1}}+\frac{2}{\sqrt{3+1}}$
$=\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}+\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$
$=\frac{\sqrt{2}+1}{(\sqrt{2})^{2}-(1)^{2}}+\frac{2 ( \sqrt{3}-1)}{(\sqrt3)^{2}-(1)^{2}}$
$=\frac{\sqrt{2}+1}{2-1}+\frac{2(\sqrt{3}-1)}{3-1}$
$=\frac{\sqrt{2}+1}{1}+\frac{2(\sqrt{3}-1)}{2}$
$=\sqrt{2}+1+\sqrt{3}-1$
$=\sqrt{2}+\sqrt{3}$
=R.H.S
L.H.S=R.H.S
Question 11
Simplify:
$\frac{6 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$
Sol :
$=\frac{6 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$
$=\frac{6\sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}-\frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}+\frac{2\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}$
Rationalising the denominator
$=\frac{6 \sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3})^{2}-(\sqrt{6})^{2}}-\frac{4 \sqrt{3}(\sqrt{6} -\sqrt{2})}{(\sqrt{6})^2-(\sqrt2)^2}+\frac{2\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^2-(\sqrt{3})^2}$
$=\frac{6 \sqrt{2}(\sqrt{3}-\sqrt{6})}{3-6}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{2 \sqrt{6}(\sqrt2-\sqrt{3})}{2-3}$
$=\frac{6 \sqrt{2}(\sqrt{3}-56)}{-3}-\frac{4 \sqrt{3}(\sqrt6-\sqrt{2})}{4}+\frac{2 \sqrt{6}(\sqrt{2}-\sqrt{3})}{-1}$
$=-2 \sqrt{2}(\sqrt{3}-\sqrt{6})-\sqrt{3}(\sqrt{6}-\sqrt{2})-2 \sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)$
$=-2 \sqrt{6}+2 \sqrt{12}-\sqrt{18}+\sqrt{6}-2\sqrt{12}+ 2\sqrt{18}$
$=-2 \sqrt{6}+2 \times 2 \sqrt{3}-3 \sqrt{2}+\sqrt{6}-2 \times 2 \sqrt{3}+2 \times \sqrt{3}+2 \times 3 \sqrt{2}$
$=-2 \sqrt{6}+4 \sqrt{3}-3 \sqrt{2}+\sqrt{6}-4 \sqrt{3}+6\sqrt{2}$
$=-2\sqrt6+\sqrt{6}+4 \sqrt{3}-4 \sqrt{3}-3 \sqrt{2}+6 \sqrt{2}$
$=-\sqrt{6}+3 \sqrt{2}$
$=3 \sqrt{2}-\sqrt{6}$
$=\sqrt{9 \times 2}-\sqrt{6}$
$=\sqrt{18}-\sqrt{6}$
Question 12
Simplify:
(i) $\frac{6}{2 \sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}} - \frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}$
(ii) $\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt6+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3\sqrt2}$
Sol :
(i) $\frac{6}{2 \sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}} - \frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}$
$\frac{6(2 \sqrt{3}+\sqrt{6})}{(2 \sqrt{3}-\sqrt{6})(2\sqrt{3}+\sqrt{6})}$
$=\frac{6(2 \sqrt{3}+\sqrt{6})}{(2 \sqrt{3})^{2}-(\sqrt{6})^{2}}$
$=\frac{6(2 \sqrt{3}+\sqrt{6})}{12-6}$
$=\frac{6(2 \sqrt{3}+\sqrt{6})}{6}$
$=2 \sqrt{3}+\sqrt{6}$...(i)
$\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$
$=\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$
$=\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2}$
$=\frac{\sqrt{18}-\sqrt{12}}{1}$
$=3 \sqrt{2}-2 \sqrt{3}$...(ii)
$\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}$
$=\frac{4 \sqrt{3}(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}$
$=\frac{4 \sqrt{3}(\sqrt{6}+\sqrt{2})}{6-2}$
$=\frac{4 \sqrt{3}(\sqrt{6}+\sqrt{2})}{4}$
$=\sqrt{3}(\sqrt{6}+\sqrt{12})$
$=\sqrt{18}+\sqrt{6}$
$=3 \sqrt{2}+\sqrt{6}$...(iii)
From (i) , (ii) , (iii)
$=(2 \sqrt{3}+\sqrt{6})+(3 \sqrt{2}-2 \sqrt{3})-(3 \sqrt{2}+\sqrt{6})$
$=2 \sqrt{3}+\sqrt{6}+3 \sqrt{2}-2 \sqrt{3}-3 \sqrt{2}-\sqrt{6}$
=0
(ii) $\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt6+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3\sqrt2}$
$=\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}$
$=\frac{7 \sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{10-3}$
$=\frac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{10-3}$
$=\frac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{7}$
$=\sqrt{3}(\sqrt{10-\sqrt{3})}$
$=\sqrt{30}-3$...(i)
$\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{2 \sqrt{5}(\sqrt{6}-\sqrt{5})}{(\sqrt{6}+\sqrt{5)}(\sqrt{6}-\sqrt{5})}$
$=\frac{2 \sqrt{5}(\sqrt{6}-\sqrt{5})}{6-5}$
$=\frac{2 \sqrt{5}(\sqrt{6}-\sqrt{5})}{1}$
$=2 \sqrt{30}-2 \times 5$
$=2 \sqrt{30}-10$...(ii)
$=\frac{3 \sqrt{2}}{\sqrt{15}+3\sqrt{2}}$
$=\frac{3 \sqrt{2}(15-3 \sqrt{2})}{(\sqrt{15}+3 \sqrt{2})(\sqrt{15}-3 \sqrt{2})}$
$=\frac{3 \sqrt{2}(\sqrt{15}-3 \sqrt{2})}{15-18}$
$=\frac{3 \sqrt{2}(\sqrt{15}-3 \sqrt{2})}{-3}$
$=-\sqrt{2}(\sqrt{15}-3 \sqrt{2})$
$=-\sqrt{30}+3 \times 2$
$=-\sqrt{30}+6$....(iii)
From (i), (ii) and (iii)
$=(\sqrt{30}-3)-(2 \sqrt{30}-10)-(-\sqrt{30}+6)$
$=\sqrt{30}-3-2 \sqrt{30}+10+\sqrt{30}-6$
=10-9
=1
Question 13
If x=2+√3 , find the value of $x^2+\frac{1}{x^2}$
Sol :
⇒x=2+√3
⇒$\frac{1}{x}=\frac{1}{2+\sqrt{3}}$
$=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}$
$=\frac{2-\sqrt{3}}{4-3}$
$=\frac{2-\sqrt{3}}{1}$
$=2-\sqrt{3}$
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}$
=4
Squaring both side
$=\left(x+\frac{1}{x}\right)^{2}=(4)^{2}$
$=x^{2}+\frac{1}{x^{2}}+2=16$
$=x^{2}+\frac{1}{x^{2}}=16-2$
=14
Question 14
If $x=\sqrt{2}+1$ , find the value of $x^2+\frac{1}{x^2}$
Sol :
⇒x=√x+1
⇒$\frac{1}{x}=\frac{1}{\sqrt{x}+1}$
$=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}$
$=\frac{\sqrt{2}-1}{(\sqrt{2})^2-1^2)}$
$=\frac{\sqrt{2}-1}{(2-1)}$
=√2-1
$x+\frac{1}{x}=\sqrt{2}+1+\sqrt{2}-1$
Squaring both sides
$\left[x+\frac{1}{x}\right]^{2}=(2 \sqrt{2})^{2}$
$=x^{2}+\frac{1}{x^{2}}+2$
=4×2=8
$=x^{2}+\frac{1}{x}=8-2$
=6
Question 15
If $x=\frac{5-\sqrt{21}}{2}$ , find the value of
(i) $x+\frac{1}{x}$ and (ii) $x^2+\frac{1}{x^2}$
Sol :
(i) $x=\frac{5-\sqrt{21}}{2}$
$=\frac{1}{x}=\frac{2}{5-\sqrt{21}}$
$=\frac{2(5+\sqrt{21})}{(5-\sqrt{21})(5+\sqrt{21})}$
$=\frac{2(5+\sqrt{2} 1)}{25-21}$
$=\frac{2(5+\sqrt{21})}{4}$
$=\frac{5+\sqrt{2} 1}{2}$
$x+\frac{1}{x}=\frac{5-\sqrt{2} 1}{2}+\frac{5+\sqrt{21}}{2}$
$=\frac{5-\sqrt{21}+5+\sqrt{21}}{2}$
$=\frac{10}{2}$
=5
$=x+\frac{1}{x}=5$
Squaring both sides
$\left(x+\frac{1}{x}\right)^{2}=(5)^{2}$
$=x^{2}+\frac{1}{x^{2}}+2$
=25
$=x^{2}+\frac{1}{x^{2}}$
=25-2
=23
$x+\frac{1}{x}=5$
$x^{2}+\frac{1}{x^{2}}=23$
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