Selected Problems on Surds
Problem 1 If \( x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} \) and \( x y=1 \), then show that
$\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}=\frac{12}{11}$
Solution :
$y=\frac{1}{x}=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$
$x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}=\frac{(\sqrt{7}+\sqrt{3})^{2}}{7-3}$
$\Rightarrow x^{2}=\frac{(7+3+2 \cdot \sqrt{21})^{2}}{4^{2}}$
$\Rightarrow x^{2}=\frac{4(5+\sqrt{21})^{2}}{4^{2}}=\frac{(5+\sqrt{21})^{2}}{4}=\frac{25+21+10 \cdot \sqrt{21}}{4}$
$\Rightarrow x^2=\frac{23+5\cdot\sqrt{21}}{2}$
Simiilarly, $\quad y^{2}=\frac{(5-\sqrt{21})^{2}}{4}=\frac{23-5 \sqrt{21}}{2}$
\Therefore,
$x^{2}+y^{2}=\frac{23+5 \sqrt{21}}{2}+\frac{23-5 \sqrt{21}}{2}=\frac{46}{2}=\frac{23}{1}$
$\frac{x^{2}+y^{2}+1}{x^{2}+y^{2}-1}=\frac{23+1}{23-1}=\frac{24}{22}$ [ By Componendo-Dividendo]
$\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}=\frac{12}{11} \quad$ (Proved)
Problem 2 If \( a=\frac{\sqrt{5}+1}{\sqrt{5}-1} \) and \( b=\frac{\sqrt{5}-1}{\sqrt{5}+1} \), then calculate the following:
(i) $\frac{a^{2}+a b+b^{2}}{a^{2}-a b+b^{2}}$
(ii) $\frac{(a-b)^{3}}{(a+b)^{3}}$
(iii) $\frac{3 a^{2}+5 a b+3 b^{2}}{3 a^{2}-5 a b+3 b^{2}}$
(iv) $\frac{a^3 + b^3}{a^3 - b^3}$
Solution:
(i)
$a^{2}=\left(\frac{\sqrt{5}+1}{\sqrt{5}-1}\right)^{2}=\frac{6+2 \sqrt{6}}{6-2 \sqrt{5}}=\frac{3+\sqrt{5}}{3-\sqrt{5}}$
$b^{2}=\left(\frac{\sqrt{5}-1}{\sqrt{5}+1}\right)^{2}=\frac{3-\sqrt{5}}{3+\sqrt{5}}$
Therefore, $\quad a^{2}+b^{2}=\frac{3+\sqrt{5}}{3-\sqrt{5}}+\frac{3-\sqrt{5}}{3+\sqrt{5}}$
$\Rightarrow \quad a^{2}+b^{2}=\frac{(3+\sqrt{5})^{2}+(3-\sqrt{5})^{2}}{9-5}=\frac{2 \times(9+5)}{4}$
$\Rightarrow \quad a^{2}+b^{2}=7$
$\Rightarrow \quad \frac{a^{2}+b^{2}+1}{a^{2}+b^{2}-1}=\frac{7+1}{7-1}$
$\Rightarrow \quad \frac{a^{2}+a b+b^{2}}{a^{2}-a b+b^{2}}=\frac{8}{6}=\frac{4}{3}$
(ii)
$\frac{a}{b} =\frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}-1} =\frac{(\sqrt{5}+1)^{2}}{(\sqrt{5}-1)^{2}}=\frac{6+2 \sqrt{5}}{6-2 \sqrt{5}}=\frac{3+\sqrt{5}}{3-\sqrt{5}}$
$\Rightarrow \frac{a-b}{a+b} =\frac{(3+\sqrt{5})-(3-\sqrt{5})}{(3+\sqrt{5})+(3-\sqrt{5})}=\frac{2 \sqrt{5}}{2.3}$
$\Rightarrow \frac{(a-b)^{3}}{(a+b)^{3}} =\frac{5 \sqrt{5}}{27}$
(iii)
$a^{2}+b^{2}=7$
$\Rightarrow 3 a^{2}+3 b^{2}=21$
$\Rightarrow \frac{3 a^{2}+3 b^{2}}{5}=\frac{21}{5}$
$\Rightarrow \frac{3 a^{2}+3 b^{2}}{5ab}=\frac{21}{5}$ [Since, $ab = 1$]
$\Rightarrow \frac{3a^{2}+5ab+3b^{2}}{3a^{2} - 5ab + 3b^{2}} = \frac{21 +5}{21-5} = \frac{26}{16} =\frac{13}{8}$
(iv)
$\frac{a}{b}=\frac{3+\sqrt{5}}{3-\sqrt{5}}$
$\Rightarrow \frac{a^{3}}{b^{3}}=\frac{(3+\sqrt{5})^{3}}{(3-\sqrt{5})^{3}}=\frac{27+27 \sqrt{5}+45+5 \sqrt{5}}{27-27 \sqrt{5}+45-5 \sqrt{5}}$
$\Rightarrow \frac{a^{3}}{b^{3}}=\frac{72+32 \sqrt{5}}{72-32 \sqrt{5}}$
$\Rightarrow \frac{a^{3}+b^{3}}{a^{3}-b^{3}}=\frac{72+32 \sqrt{5}+72-32 \sqrt{5}}{72+32 \sqrt{5}-\sqrt{2}+32 \sqrt{5}}$
$\Rightarrow \frac{a^{3}+b^{3}}{a^{3}-b^{3}}=\frac{2 \cdot 72 \cdot 9}{2 \cdot 32 \sqrt{5}}=\frac{9}{4 \sqrt{5}}=\frac{9 \sqrt{5}}{20}$
Problem 3. If \( \frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}} = 14 \), then find out the possible values of x.
Solution:
Let, $\quad p=\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}=\frac{\left(x+\sqrt{x^{2}-1}\right)\left(x+\sqrt{x^{2}-1}\right)}{x^{2}-\left(x^{2}-1\right)}$
$\Rightarrow p=\frac{x^{2}+x^{2}-1+2 x \sqrt{x^{2}-1}}{1}$
$\Rightarrow p=2 x^{2}-1+2 x \sqrt{x^{2}-1}$
Therefore, \( \frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}=\frac{1}{p}=2 x^{2}-1-2 x \sqrt{x^{2}-1} \)
Hence, $L . H . S =p+\frac{1}{p}$
$ =2 x^{2}-1+2 x \sqrt{x^{2}-1}+2 x^{2}-1-2 x \sqrt{x^{2}-1}$
$ =4 x^{2}-2$
Now, if $p+\frac{1}{p}=14$
$\Rightarrow 4 x^{2}-2=14$
$\Rightarrow 4 x^{2}=16$
$\Rightarrow x^{2}=4$
$\Rightarrow x= \pm 2$
Problem 4. Find the value of
\( \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} \)
Solution:
$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}$
$= \frac{\sqrt{2} - \sqrt{1}}{2 - 1} + \frac{\sqrt{3} - \sqrt{2}}{3 - 2} + \frac{\sqrt{4} - \sqrt{3}}{4 - 3} + \frac{\sqrt{5} - \sqrt{4}}{5 - 4} + \frac{\sqrt{6} - \sqrt{5}}{6 - 5} + \frac{\sqrt{7} - \sqrt{6}}{7 - 6} + \frac{\sqrt{8} - \sqrt{7}}{8 - 7} + \frac{\sqrt{9} - \sqrt{8}}{9 - 8}$
$= \sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3}+ \sqrt{5} - \sqrt{4} + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + \sqrt{8} - \sqrt{7} + \sqrt{9} - \sqrt{8}$
$= \sqrt{9} - \sqrt{1}$
$=3 - 1 = 2$
