Rational & Irrational Numbers — Exercise 1(D)

Rationalize the denominator:

$\dfrac{2}{\sqrt{6}}$

Answer

(i) Rationalizing the denominator,

$\Rightarrow \dfrac{2}{\sqrt{6}} \times \dfrac{ \sqrt{6}}{\sqrt{6}} \\[1em] \Rightarrow \dfrac{2(\sqrt{6})}{(\sqrt{6})^2} \\[1em] \Rightarrow \dfrac{2\sqrt{6}}{6} \\[1em] \Rightarrow \dfrac{\sqrt{6}}{3}$

Hence, on rationalizing $\dfrac{2}{\sqrt{6}} = \dfrac{\sqrt{6}}{3}$.

Rationalize the denominator:

$\dfrac{\sqrt{2}}{2\sqrt{3}}$

Answer

Rationalizing the denominator, $\Rightarrow \dfrac{\sqrt{2}}{2\sqrt{3}} \times \dfrac{2\sqrt{3}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2(\sqrt{6})}{(2\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{2\sqrt{6}}{(4 \times 3)} \\[1em] \Rightarrow \dfrac{2\sqrt{6}}{12} \\[1em] \Rightarrow \dfrac{\sqrt{6}}{6}.$

Hence, on rationalizing = $\dfrac{\sqrt{2}}{2\sqrt{3}} = \dfrac{\sqrt{6}}{6}$.

Rationalize the denominator:

$\dfrac{1}{(3 + \sqrt{5})}$.

Answer

Rationalizing the denominator,

$\Rightarrow \dfrac{1}{(3 + \sqrt{5})} \times \dfrac{(3 - \sqrt{5})} {(3 - \sqrt{5})} \\[1em] \Rightarrow \dfrac{(3 - \sqrt{5})} {(3)^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{(3 - \sqrt{5})} {9 - 5} \\[1em] \Rightarrow \dfrac{(3 - \sqrt{5})} {4} \\[1em]$

Hence, on rationalizing $\dfrac{1}{(3 + \sqrt{5})} = \dfrac{(3 - \sqrt{5})} {4}$.

Rationalize the denominator:

$\dfrac{1}{(\sqrt{3} - 1)}$

Answer

Rationalizing on denominator,

$\Rightarrow \dfrac{1}{(\sqrt{3} - 1)} \times \dfrac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)}{3 - 1} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)}{2} \\[1em]$

Hence, on rationalizing $\dfrac{1}{(\sqrt{3} - 1)} = \dfrac{(\sqrt{3} + 1)}{2}$.

Rationalize the denominator:

$\dfrac{1}{(4 + 2\sqrt{3})}$

Answer

Rationalizing the denominator,

$\Rightarrow \dfrac{1}{(4 + 2\sqrt{3})} \times \dfrac{(4 - 2\sqrt{3})} {(4 - 2\sqrt{3})} \\[1em] \Rightarrow \dfrac{(4 - 2\sqrt{3})} {(4)^2 - (2\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(4 - 2\sqrt{3})} {16 - 12} \\[1em] \Rightarrow \dfrac{2(2 - \sqrt{3})} {4} \\[1em] \Rightarrow \dfrac{(2 - \sqrt{3})} {2} \\[1em]$

Hence, on rationalizing $\dfrac{1}{(4 + 2\sqrt{3})} = \dfrac{(2 - \sqrt{3})} {2}$.

Rationalize the denominator:

$\dfrac{1}{(\sqrt{6} - \sqrt{3})}$

Answer

Rationalizing the denominator,

$\Rightarrow \dfrac{1}{(\sqrt{6} - \sqrt{3})} \times \dfrac{(\sqrt{6} + \sqrt{3})}{(\sqrt{6} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{(\sqrt{6})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{6 - 3} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{3} \\[1em]$

Hence, on rationalizing $\dfrac{1}{(\sqrt{6} - \sqrt{3})} = \dfrac{(\sqrt{6} + \sqrt{3})}{3}$.

Rationalize the denominator:

$\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}$.

Answer

Rationalizing the denominator,

$\Rightarrow \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} \\[1em] \Rightarrow \dfrac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow \dfrac{(\sqrt{3})^2 + (1)^2 - 2 \times \sqrt{3} \times 1 }{3 - 1} \\[1em] \Rightarrow \dfrac{3 + 1 - 2\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{(4 - 2\sqrt{3})}{2} \\[1em] \Rightarrow \dfrac{2(2 - \sqrt{3})}{2} \\[1em] \Rightarrow (2 - \sqrt{3})$

Hence, on rationalizing = $\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} = (2 - \sqrt{3})$.

Rationalize the denominator:

$\dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}$

Answer

Rationalizing the denominator,

$\Rightarrow \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \dfrac{(3 - 2\sqrt{2})^2} {(3)^2 - ( 2\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{(3)^2 + ( 2\sqrt{2})^2 -2 \times 3 \times 2\sqrt{2} } {9-8} \\[1em] \Rightarrow \dfrac{(9 + 8 - 12\sqrt{2})} {1} \\[1em] \Rightarrow (17 - 12\sqrt{2}) \\[1em]$

Hence, on rationalizing $\dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = (17 - 12\sqrt{2})$.

Rationalize the denominator:

$\dfrac{1}{(2\sqrt{5} - \sqrt{3})}$.

Answer

Rationalizing the denominator,

$\Rightarrow \dfrac{1}{(2\sqrt{5} - \sqrt{3})} \times \dfrac{(2\sqrt{5} + \sqrt{3})} {(2\sqrt{5} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {(2\sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {20 - 3} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {17} \\[1em]$

Hence, on rationalizing $\dfrac{1}{(2\sqrt{5} - \sqrt{3})} = \dfrac{(2\sqrt{5} + \sqrt{3})}{17}$.

Rationalize the denominator:

$\dfrac{1}{(1 + \sqrt{5} + \sqrt{3})}$.

Answer

Rationalizing the denominator,

$\Rightarrow \dfrac{1}{(1 + \sqrt{5} + \sqrt{3})} \times \dfrac{(1 + \sqrt{5} - \sqrt{3})} {(1 + \sqrt{5} - \sqrt{3})} \\[1em] \Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{(1 + \sqrt{5} + \sqrt{3}) \times (1 + \sqrt{5} - \sqrt{3})} \\[1em] \Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{(1 + \sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{1 + 5 + 2\sqrt{5} - 3} \\[1em] \Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{(3 + 2\sqrt{5})}$

Rationalizing the denominator again,

$\Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{(3 + 2\sqrt{5})} \times \dfrac {(3 - 2\sqrt{5})}{(3 - 2\sqrt{5})} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{5} + 3\sqrt{5} - 10 - 3\sqrt{3} + 2\sqrt{15}}{(3)^2 - (2\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{-7 + \sqrt{5} - 3\sqrt{3} + 2\sqrt{15}}{9 - 20} \\[1em] \Rightarrow \dfrac{-(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})}{-11} \\[1em] \Rightarrow \dfrac{(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})}{11}$

Hence, on rationalizing $\dfrac{1}{(1 + \sqrt{5} + \sqrt{3})} = \dfrac{(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})} {11}$.

Rationalize the denominator:

$\dfrac{\sqrt{2}}{(\sqrt{2} + \sqrt{3} - \sqrt{5})}$.

Answer

Rationalizing the denomaintor,

$\Rightarrow \dfrac{\sqrt{2}}{(\sqrt{2} + \sqrt{3} - \sqrt{5})} \times \dfrac{(\sqrt{2} + \sqrt{3} + \sqrt{5})}{(\sqrt{2} + \sqrt{3} + \sqrt{5})} \\[1em] \Rightarrow \dfrac{ \sqrt{2} \times (\sqrt{2}+ \sqrt{3} + \sqrt{5})} {(\sqrt{2}+ \sqrt{3} - \sqrt{5}) \times (\sqrt{2} + \sqrt{3} + \sqrt{5})} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{6} + \sqrt{10})} {(2 + \sqrt{6} + \sqrt{10} + \sqrt{6} + 3 + \sqrt{15} - \sqrt{10} - \sqrt{15} - 5)} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{6} + \sqrt{10})} {(2\sqrt{6})}$

Rationalizing again,

$\Rightarrow \dfrac{(2 + \sqrt{6} + \sqrt{10})} {(2\sqrt{6})} \times \dfrac{\sqrt{6}}{\sqrt{6}} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{6} + \sqrt{10}) \times \sqrt{6}} {(2\sqrt{6})\times \sqrt{6}} \\[1em] \Rightarrow \dfrac{(2\sqrt{6} + 6 + \sqrt{60})} {12} \\[1em] \Rightarrow \dfrac{(2\sqrt{6} + 6 + \sqrt{15 \times 4})}{12} \\[1em] \Rightarrow \dfrac{2\sqrt{6} + 6 + 2\sqrt{15}}{12} \\[1em] \Rightarrow \dfrac{2(\sqrt{6} + 3 + \sqrt{15})}{12} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + 3 + \sqrt{15})}{6}.$

Hence, on rationalizing $\dfrac{(\sqrt{6} + 3 + \sqrt{15})}{6} = \dfrac{(\sqrt{6} + 3 + \sqrt{15})}{6}$.

If $\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = a + b\sqrt{3}$ ,find the values of 'a' and 'b'.

Answer

Given,

Equation : $\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = a + b\sqrt{3}$

Rationalizing the denominator of L.H.S. of the above equation :

$\Rightarrow \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow \dfrac{(\sqrt{3})^2 + 1^2 + 2 \times \sqrt{3} \times 1}{3 - 1} \\[1em] \Rightarrow \dfrac{3 + 1 + 2\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{4 + 2\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{2(2 + \sqrt{3})}{2} \\[1em] \Rightarrow 2 + \sqrt{3}.$

Comparing $2 + \sqrt{3} \text{ with } a + b\sqrt{3}$, we get :

a = 2 and b = 1.

Hence, a = 2 and b = 1.

If $\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b\sqrt{2}$, find the values of 'a' and 'b'.

Answer

Given,

Equation : $\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b\sqrt{2}$

Rationalizing L.H.S. of the above equation :

$\Rightarrow \dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}} \\[1em] \Rightarrow \dfrac{(3 + \sqrt{2})^2}{(3)^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{(3)^2 + (\sqrt{2})^2 + 2 \times 3 \times \sqrt{2}}{9 - 2} \\[1em] \Rightarrow \dfrac{9 + 2 + 6\sqrt{2}}{7} \\[1em] \Rightarrow \dfrac{11 + 6\sqrt{2}}{7} \\[1em] \Rightarrow \dfrac{11}{7} + \dfrac{6\sqrt{2}}{7}$

Comparing $\dfrac{11}{7} + \dfrac{6}{7}\sqrt{2} \text{ with } a + b\sqrt{2}$, we get :

$a = \dfrac{11}{7} \text{ and } b = \dfrac{6}{7}.$

Hence, $a = \dfrac{11}{7} \text{ and } b = \dfrac{6}{7}$.

If $\dfrac{5 - \sqrt{6}}{5 + \sqrt{6}} = a - b\sqrt{6}$, find the values of 'a' and 'b'.

Answer

Given,

Equation : $\dfrac{5 - \sqrt{6}}{5 + \sqrt{6}} = a - b\sqrt{6}$

Rationalizing the denomiantor of L.H.S. of the above equation :

$\Rightarrow \dfrac{5 - \sqrt{6}}{5 + \sqrt{6}} \times \dfrac{5 - \sqrt{6}}{5 - \sqrt{6}} \\[1em] \Rightarrow \dfrac{(5 - \sqrt{6})^2}{(5)^2 - (\sqrt{6})^2} \\[1em] \Rightarrow \dfrac{(5)^2 + (\sqrt{6})^2 - 2 \times 5 \times \sqrt{6}}{25 - 6} \\[1em] \Rightarrow \dfrac{25 + 6 - 10\sqrt{6}}{19} \\[1em] \Rightarrow \dfrac{31 - 10\sqrt{6}}{19} \\[1em] \Rightarrow \dfrac{31}{19} - \dfrac{10\sqrt{6}}{19}$

Comparing $\dfrac{31}{19} - \dfrac{10}{19}\sqrt{6} \text{ with } a - b\sqrt{6}$, we get :

$a = \dfrac{31}{19} \text{ and } b = \dfrac{10}{19}.$

Hence, $a = \dfrac{31}{19} \text{ and } b = \dfrac{10}{19}$.

If $\dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = a - b\sqrt{3}$, find the values of 'a' and 'b'.

Answer

Given,

Equation : $\dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = a - b\sqrt{3}$

Rationalizing the denominator of L.H.S. of the above equation :

$\Rightarrow \dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} \times \dfrac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} \\[1em] \Rightarrow \dfrac{(5 + 2\sqrt{3}) \times (7 - 4\sqrt{3})}{(7)^2 - (4\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{35 - 20\sqrt{3} + 14\sqrt{3} - 8 \times 3}{49 - 48} \\[1em] \Rightarrow \dfrac{35 - 6\sqrt{3} - 24}{1} \\[1em] \Rightarrow 11 - 6\sqrt{3} \\[1em]$

Comparing, $11 - 6\sqrt{3} \text{ with } a - b\sqrt{3}$, we get :

a = 11 and b = 6.

Hence, a = 11 and b = 6.

Simplify : $\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

Answer

Given,

Equation : $\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

Simplifying the above equation :

$\Rightarrow \dfrac{(\sqrt{5} + \sqrt{3})^2 + (\sqrt{5} - \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{(\sqrt{5})^2 + (\sqrt{3})^2 + 2 \times \sqrt{5} \times \sqrt{3} + (\sqrt{5})^2 + (\sqrt{3})^2 - 2 \times \sqrt{5} \times \sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{5 + 3 + 2\sqrt{15} + 5 + 3 - 2\sqrt{15}}{5 - 3} \\[1em] \Rightarrow \dfrac{16}{2} \\[1em] \Rightarrow 8$

Hence, $\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}+\dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ = 8.

Simplify : $\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}$

Answer

Given,

Equation : $\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}$

Simplifying the above equation :

$\Rightarrow \dfrac{(7 + 3\sqrt{5}) \times (3 - \sqrt{5}) - (7 - 3\sqrt{5}) \times (3 + \sqrt{5})}{(3 + \sqrt{5}) \times (3 - \sqrt{5})} \\[1em] \Rightarrow \dfrac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15)-(21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{3^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{21 - 7\sqrt{5} + 9\sqrt{5} - 15 - 21 - 7\sqrt{5} + 9\sqrt{5} + 15}{9 - 5} \\[1em] \Rightarrow \dfrac{4\sqrt{5}}{4} \\[1em] \Rightarrow \sqrt{5}$

Hence, $\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}}- \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = \sqrt{5}$.

Show that : $\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} = 5$

Answer

Given,

Equation : $\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})}$

Simplifying L.H.S. of the above equation :

$\Rightarrow \dfrac{1}{(3 - \sqrt{8})} \times \dfrac{(3 + \sqrt{8})}{(3 + \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})} \times \dfrac{(\sqrt{7} + \sqrt{6})}{(\sqrt{7} + \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} \times \dfrac{(\sqrt{5} + 2)}{(\sqrt{5} + 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} \times \dfrac{(\sqrt{8} + \sqrt{7})}{(\sqrt{8} + \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} \times \dfrac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6} + \sqrt{5})} \\[1em] \Rightarrow \dfrac{3 + \sqrt{8}}{3^2 - (\sqrt{8})^2} + \dfrac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} + \dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} - \dfrac{\sqrt{8} + \sqrt{7}}{(\sqrt{8})^2 - (\sqrt{7})^2} - \dfrac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{3 + \sqrt{8}}{9 - 8} + \dfrac{\sqrt{7} + \sqrt{6}}{7 - 6} + \dfrac{\sqrt{5} + 2}{5 - 4} - \dfrac{\sqrt{8} + \sqrt{7}}{8 - 7} - \dfrac{\sqrt{6} + \sqrt{5}}{6 - 5} \\[1em] \Rightarrow 3 + \sqrt{8} + \sqrt{7} + \sqrt{6} + \sqrt{5} + 2 - (\sqrt{8} + \sqrt{7}) - (\sqrt{6} + \sqrt{5}) \\[1em] \Rightarrow 3 + 2 + \sqrt{8} - \sqrt{8} + \sqrt{7} - \sqrt{7} + \sqrt{6} - \sqrt{6} + \sqrt{5} - \sqrt{5} \\[1em] \Rightarrow 5$

Hence, proved that

$\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} = 5$.

If x = $(3 + \sqrt{8})$, find the values of $\Big(x^2 + \dfrac{1}{x^2}\Big)$.

Answer

Given,

x = $(3 + \sqrt{8})$

$\therefore \dfrac{1}{x} = \dfrac{1}{(3 + \sqrt{8})}$

Rationalizing,

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{(3 + \sqrt{8})} \times \dfrac{(3 - \sqrt{8})}{(3 - \sqrt{8})} \\[1em] = \dfrac{3 - \sqrt{8}}{3^2 - (\sqrt{8})^2} \\[1em] = \dfrac{3 - \sqrt{8}}{9 - 8} \\[1em] = \dfrac{3 - \sqrt{8}}{1} \\[1em] = 3 - \sqrt{8}.$

By formula,

$x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2$

Substituting values we get :

$\Rightarrow x^2 + \dfrac{1}{x^2} = (3 + \sqrt{8} + 3 - \sqrt{8})^2 - 2 \\[1em] = 6^2 - 2 \\[1em] = 36 - 2 \\[1em] = 34.$

Hence, $x^2 + \dfrac{1}{x^2}$ = 34.

If x = $(4 - \sqrt{15})$, find the values of $\Big(x^2 + \dfrac{1}{x^2}\Big)$.

Answer

Given,

x = $(4 - \sqrt{15})$

$\therefore \dfrac{1}{x} = \dfrac{1}{(4 - \sqrt{15})}$

Rationalizing,

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{(4 - \sqrt{15})} \times \dfrac{(4 + \sqrt{15})}{(4 + \sqrt{15})} \\[1em] = \dfrac{4 + \sqrt{15}}{4^2 - (\sqrt{15})^2} \\[1em] = \dfrac{4 + \sqrt{15}}{16 - 15} \\[1em] = \dfrac{4 + \sqrt{15}}{1} \\[1em] = 4 + \sqrt{15}.$

By formula,

$x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2$

Substituting values we get :

$\Rightarrow x^2 + \dfrac{1}{x^2} = (4 - \sqrt{15} + 4 + \sqrt{15})^2 - 2 \\[1em] = 8^2 - 2 \\[1em] = 64 - 2 \\[1em] = 62.$

Hence, $x^2 + \dfrac{1}{x^2}$ = 62.