(i) Given x = 7 + 3 5 7 − 3 5 \dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} 7 − 3 5 7 + 3 5
Rationalising the denominator,
7 + 3 5 7 − 3 5 = 7 + 3 5 7 − 3 5 × 7 + 3 5 7 + 3 5 = ( 7 + 3 5 ) 2 ( 7 ) 2 − ( 3 5 ) 2 = 7 2 + 2 × 7 × 3 5 + ( 3 5 ) 2 49 − 45 = 49 + 42 5 + 45 4 = 94 + 42 5 4 = 47 + 21 5 2 ∴ x = 47 + 21 5 2 ⇒ 1 x = 2 47 + 21 5 \dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} = \dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} × \dfrac{7 + 3\sqrt{5}}{7 + 3\sqrt{5}} \\[1.5em] = \dfrac{({7 + 3\sqrt{5}})^2}{(7)^2 - (3\sqrt{5})^2} \\[1.5em] = \dfrac{7^2 + 2 × 7 × 3\sqrt{5} + (3\sqrt{5})^2}{49 - 45} = \dfrac{49 + 42\sqrt{5} + 45}{4} = \dfrac{94 + 42\sqrt{5}}{4} \\[1.5em] = \dfrac{47 + 21\sqrt{5}}{2} \\[1.5em] \therefore x = \dfrac{47 + 21\sqrt{5}}{2} \\[1.5em] \Rightarrow \dfrac{1}{x} = \dfrac{2}{47 + 21\sqrt{5}} \\[1.5em] 7 − 3 5 7 + 3 5 = 7 − 3 5 7 + 3 5 × 7 + 3 5 7 + 3 5 = ( 7 ) 2 − ( 3 5 ) 2 ( 7 + 3 5 ) 2 = 49 − 45 7 2 + 2 × 7 × 3 5 + ( 3 5 ) 2 = 4 49 + 42 5 + 45 = 4 94 + 42 5 = 2 47 + 21 5 ∴ x = 2 47 + 21 5 ⇒ x 1 = 47 + 21 5 2
Rationalising denominator of 1 x \dfrac{1}{x} x 1
1 x = 2 47 + 21 5 × 47 − 21 5 47 − 21 5 = 2 ( 47 − 21 5 ) ( 47 ) 2 − ( 21 5 ) 2 = 2 ( 47 − 21 5 ) 2209 − 2205 = 2 ( 47 − 21 5 ) 4 = ( 47 − 21 5 ) 2 \dfrac{1}{x} = \dfrac{2}{47 + 21\sqrt{5}} × \dfrac{47 - 21\sqrt{5}}{47 - 21\sqrt{5}} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {(47)^2 - (21\sqrt{5})^2} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {2209 - 2205} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {4} \\[1.5em] = \dfrac{(47 - 21\sqrt{5})} {2} \\[1.5em] x 1 = 47 + 21 5 2 × 47 − 21 5 47 − 21 5 = ( 47 ) 2 − ( 21 5 ) 2 2 ( 47 − 21 5 ) = 2209 − 2205 2 ( 47 − 21 5 ) = 4 2 ( 47 − 21 5 ) = 2 ( 47 − 21 5 )
Now,
( x + 1 x ) = 47 + 21 5 2 + ( 47 − 21 5 ) 2 = 47 + 21 5 + 47 − 21 5 2 = 94 2 = 47 ∴ ( x + 1 x ) = 47 ….(i) {\Big(x + \dfrac{1}{x}\Big)} = \dfrac{47 + 21\sqrt{5}}{2} + \dfrac{(47 - 21\sqrt{5})}{2} \\[1.5em] = \dfrac{47 + 21\sqrt{5} + 47 -21\sqrt{5}}{2} \\[1.5em] = \dfrac{94}{2} = 47 \\[1.5em] \therefore {\Big(x + \dfrac{1}{x}\Big)} = 47 \qquad \text{….(i)} ( x + x 1 ) = 2 47 + 21 5 + 2 ( 47 − 21 5 ) = 2 47 + 21 5 + 47 − 21 5 = 2 94 = 47 ∴ ( x + x 1 ) = 47 ….(i)
We know that ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 {\Big(x + \dfrac{1}{x}\Big)}^2 = x^2 + \dfrac{1}{x^2} + 2 ( x + x 1 ) 2 = x 2 + x 2 1 + 2
⇒ x 2 + 1 x 2 = ( x + 1 x ) 2 − 2 ⇒ x 2 + 1 x 2 = ( 47 ) 2 − 2 . . . . . using(i) ⇒ x 2 + 1 x 2 = 2209 − 2 = 2207 x 2 + 1 x 2 = 2207 \Rightarrow x^2 + \dfrac{1}{x^2} = {\Big(x + \dfrac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = (47)^2 - 2 \qquad ….. \text{using(i)} \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2209 - 2 = 2207 \\[1.5em] \bold{x^2 + \dfrac{1}{x^2} = 2207} \\[1.5em] ⇒ x 2 + x 2 1 = ( x + x 1 ) 2 − 2 ⇒ x 2 + x 2 1 = ( 47 ) 2 − 2 ….. using(i) ⇒ x 2 + x 2 1 = 2209 − 2 = 2207 x 2 + x 2 1 = 2207
(ii) x = 5 − 2 5 + 2 and y = 5 + 2 5 − 2 ∴ x + y = 5 − 2 5 + 2 + 5 + 2 5 − 2 ⇒ ( 5 − 2 ) 2 + ( 5 + 2 ) 2 ( 5 + 2 ) ( 5 − 2 ) ⇒ ( 5 ) 2 − 2 × 2 × 5 + ( 2 ) 2 + ( 5 ) 2 + 2 × 2 × 5 + ( 2 ) 2 ( 5 ) 2 − ( 2 ) 2 ⇒ 5 − 2 10 + 2 + 5 + 2 10 + 2 5 − 2 = 14 3 ⇒ x + y = 14 3 ….(i) Also x y = 5 − 2 5 + 2 × 5 + 2 5 − 2 = 1 ….(ii) \text{(ii) } x = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} + \sqrt{2}} \text{ and y} = \dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \therefore x+y = \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} +\dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \Rightarrow\dfrac{(\sqrt{5} - \sqrt{2})^2 + (\sqrt{5} + \sqrt{2})^2}{(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})} \\[1.5em] \Rightarrow\dfrac{(\sqrt{5})^2 - 2 × \sqrt{2} × \sqrt{5} + (\sqrt{2})^2+ (\sqrt{5})^2 + 2 × \sqrt{2} × \sqrt{5} + (\sqrt{2})^2 }{(\sqrt{5})^2 - (\sqrt{2})^2 } \\[1.5em] \Rightarrow\dfrac{5 - 2\sqrt{10} + 2 + 5 + 2\sqrt{10} + 2}{5 - 2} = \dfrac{14}{3} \\[1.5em] \Rightarrow x + y = \dfrac{14}{3} \qquad \text{….(i)} \\[1.5em] \text{Also } xy = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} + \sqrt{2}} ×\dfrac{\sqrt{5} + \sqrt{2} }{\sqrt{5} - \sqrt{2}} = 1 \qquad \text{….(ii)} \\[1.5em] (ii) x = 5 + 2 5 − 2 and y = 5 − 2 5 + 2 ∴ x + y = 5 + 2 5 − 2 + 5 − 2 5 + 2 ⇒ ( 5 + 2 ) ( 5 − 2 ) ( 5 − 2 ) 2 + ( 5 + 2 ) 2 ⇒ ( 5 ) 2 − ( 2 ) 2 ( 5 ) 2 − 2 × 2 × 5 + ( 2 ) 2 + ( 5 ) 2 + 2 × 2 × 5 + ( 2 ) 2 ⇒ 5 − 2 5 − 2 10 + 2 + 5 + 2 10 + 2 = 3 14 ⇒ x + y = 3 14 ….(i) Also x y = 5 + 2 5 − 2 × 5 − 2 5 + 2 = 1 ….(ii)
We need to find the value of x 2 + x y + y 2 {x^2 + xy + y^2} x 2 + x y + y 2 x 2 + x y + y 2 = x 2 + y 2 + 2 x y − x y ⇒ x 2 + x y + y 2 = ( x + y ) 2 − x y ….(iii) {x^2 + xy + y^2} = x^2 + y^2 + 2xy - xy \\[1.5em] \Rightarrow {x^2 + xy + y^2} = (x+y)^2 - xy \qquad \text{….(iii)} \\[1.5em] x 2 + x y + y 2 = x 2 + y 2 + 2 x y − x y ⇒ x 2 + x y + y 2 = ( x + y ) 2 − x y ….(iii)
Substituting the values from (i) and (ii) in (iii),
x 2 + x y + y 2 = ( 14 3 ) 2 − 1 = 196 9 − 1 = 196 − 9 9 = 187 9 ∴ x 2 + x y + y 2 = 187 9 {x^2 + xy + y^2} = \Big(\dfrac{14}{3}\Big)^2 - 1 = \dfrac{196}{9} - 1 = \dfrac{196 - 9}{9} = \dfrac{187}{9} \\[1.5em] \therefore \bold{x^2 + xy + y^2} = \bold{\dfrac{187}{9}} \\[1.5em] x 2 + x y + y 2 = ( 3 14 ) 2 − 1 = 9 196 − 1 = 9 196 − 9 = 9 187 ∴ x 2 + xy + y 2 = 9 187
(iii) x = 3 − 2 3 + 2 and y = 3 + 2 3 − 2 ∴ x + y = 3 − 2 3 + 2 + 3 + 2 3 − 2 = ( 3 − 2 ) 2 + ( 3 + 2 ) 2 ( 3 + 2 ) ( 3 − 2 ) = ( 3 ) 2 − 2 × 2 × 3 + ( 2 ) 2 + ( 3 ) 2 + 2 × 2 × 3 + ( 2 ) 2 ( 3 ) 2 − ( 2 ) 2 = 3 − 2 6 + 2 + 3 + 2 6 + 2 3 − 2 = 10 1 = 10 ∴ x + y = 10 ….(i) Also x y = 3 − 2 3 + 2 × 3 + 2 3 − 2 = 1 ….(ii) \text{(iii) } x = \dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}} \text{ and y} = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1.5em] \therefore x+y = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} +\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1.5em] = \dfrac{(\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} \\[1.5em] = \dfrac{(\sqrt{3})^2 - 2 × \sqrt{2} × \sqrt{3} + (\sqrt{2})^2+ (\sqrt{3})^2 + 2 × \sqrt{2} × \sqrt{3} + (\sqrt{2})^2 }{(\sqrt{3})^2 - (\sqrt{2})^2 } \\[1.5em] = \dfrac{3 - 2\sqrt{6} + 2 + 3 + 2\sqrt{6} + 2}{3 - 2} = \dfrac{10}{1} = 10 \\[1.5em] \therefore x + y = 10 \qquad \text{….(i)} \\[1.5em] \text{Also } xy = \dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}} ×\dfrac{\sqrt{3} + \sqrt{2} }{\sqrt{3} - \sqrt{2}} = 1 \qquad \text{….(ii)} \\[1.5em] (iii) x = 3 + 2 3 − 2 and y = 3 − 2 3 + 2 ∴ x + y = 3 + 2 3 − 2 + 3 − 2 3 + 2 = ( 3 + 2 ) ( 3 − 2 ) ( 3 − 2 ) 2 + ( 3 + 2 ) 2 = ( 3 ) 2 − ( 2 ) 2 ( 3 ) 2 − 2 × 2 × 3 + ( 2 ) 2 + ( 3 ) 2 + 2 × 2 × 3 + ( 2 ) 2 = 3 − 2 3 − 2 6 + 2 + 3 + 2 6 + 2 = 1 10 = 10 ∴ x + y = 10 ….(i) Also x y = 3 + 2 3 − 2 × 3 − 2 3 + 2 = 1 ….(ii)
We need to find the value of x 3 + y 3 {x^3 + y^3} x 3 + y 3 x 3 + y 3 = ( x + y ) 3 − 3 x y ( x + y ) ….(iii) {x^3 + y^3} = (x+y)^3 - 3xy(x + y) \qquad \text{….(iii)} \\[1.5em] x 3 + y 3 = ( x + y ) 3 − 3 x y ( x + y ) ….(iii)
Substituting the values from (i) and (ii) in (iii),
x 3 + y 3 = ( 10 ) 3 − 3 × 1 × 10 = 1000 − 30 = 970 ∴ x 3 + y 3 = 970 {x^3+ y^3} = (10)^3 - 3 × 1 × 10 \\[1.5em] = 1000 - 30 \\[1.5em] = 970 \\[1.5em] \therefore \bold{x^3+ y^3 = 970} x 3 + y 3 = ( 10 ) 3 − 3 × 1 × 10 = 1000 − 30 = 970 ∴ x 3 + y 3 = 970
