myhelper: RD Sharma 2020 solution class 9 chapter 3 Rationalisation FBQS

FBQS

Page-3.18

Question 1:

The number obtained by rationalizing the denominator of

$\frac{1}{\sqrt{7} + 2}$ is __________.

Answer 1:

$\frac{1}{\sqrt{7} + 2} Multiply and divide by (\sqrt{7} - 2), we get = \frac{1}{\sqrt{7} + 2} \times \frac{\sqrt{7} - 2}{\sqrt{7} - 2} = \frac{\sqrt{7} - 2}{{(\sqrt{7})}^{2} - 2^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) = \frac{\sqrt{7} - 2}{7 - 4} = \frac{\sqrt{7} - 2}{3}$

Hence, the number obtained by rationalizing the denominator of

$\frac{1}{\sqrt{7} + 2}$ is

$\frac{\sqrt{7} - 2}{3} .$ .

Question 2:

$\frac{1}{\sqrt{9} - \sqrt{8}} = A + B \sqrt{2}$ , then A = ____________ and B = ____________.

Answer 2:

$\frac{1}{\sqrt{9} - \sqrt{8}} Multiply and divide by (\sqrt{9} + \sqrt{8}), we get = \frac{1}{\sqrt{9} - \sqrt{8}} \times \frac{\sqrt{9} + \sqrt{8}}{\sqrt{9} + \sqrt{8}} = \frac{\sqrt{9} + \sqrt{8}}{{(\sqrt{9})}^{2} - {(\sqrt{8})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) = \frac{\sqrt{9} + \sqrt{8}}{9 - 8} = \frac{\sqrt{9} + \sqrt{8}}{1} = \sqrt{9} + \sqrt{8} = 3 + 2 \sqrt{2} Now, it is given that \frac{1}{\sqrt{9} - \sqrt{8}} = A + B \sqrt{2} \Rightarrow 3 + 2 \sqrt{2} = A + B \sqrt{2} \Rightarrow A = 3 and B = 2$

Hence, if

$\frac{1}{\sqrt{9} - \sqrt{8}} = A + B \sqrt{2}$ , then A = 3 and B = 2.

Question 3:

After rationalizing the denominator of

$\frac{7}{3 \sqrt{3} - 2 \sqrt{2}}$ , we get the denominator as __________.

Answer 3:

$\frac{7}{3 \sqrt{3} - 2 \sqrt{2}} Multiply and divide by (3 \sqrt{3} + 2 \sqrt{2}), we get = \frac{7}{3 \sqrt{3} - 2 \sqrt{2}} \times \frac{3 \sqrt{3} + 2 \sqrt{2}}{3 \sqrt{3} + 2 \sqrt{2}} = \frac{7 (3 \sqrt{3} + 2 \sqrt{2})}{{(3 \sqrt{3})}^{2} - {(2 \sqrt{2})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) = \frac{21 \sqrt{3} + 14 \sqrt{2}}{27 - 8} = \frac{21 \sqrt{3} + 14 \sqrt{2}}{19}$

Hence, after rationalizing the denominator of

$\frac{7}{3 \sqrt{3} - 2 \sqrt{2}}$ , we get the denominator as

$\frac{21 \sqrt{3} + 14 \sqrt{2}}{19} .$

Question 4:

$a = 2 + \sqrt{3}, then a - \frac{1}{a} =___________.$

Answer 4:

$Given : a = 2 + \sqrt{3} Now, \frac{1}{a} = \frac{1}{2 + \sqrt{3}} Multiply and divide RHS by (2 - \sqrt{3}), we get \Rightarrow \frac{1}{a} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} \Rightarrow \frac{1}{a} = \frac{2 - \sqrt{3}}{{(2)}^{2} - {(\sqrt{3})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow \frac{1}{a} = \frac{2 - \sqrt{3}}{4 - 3} \Rightarrow \frac{1}{a} = \frac{2 - \sqrt{3}}{1} \Rightarrow \frac{1}{a} = 2 - \sqrt{3} Thus, a - \frac{1}{a} = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2 \sqrt{3}$

Hence, if

$a = 2 + \sqrt{3}, then a - \frac{1}{a} = 2 \sqrt{3} .$

Question 5:

$a = 5 + 2 \sqrt{6}, then a + \frac{1}{a} =_________.$

Answer 5:

$Given : a = 5 + 2 \sqrt{6} Now, \frac{1}{a} = \frac{1}{5 + 2 \sqrt{6}} Multiply and divide RHS by (5 - 2 \sqrt{6}), we get \Rightarrow \frac{1}{a} = \frac{1}{5 + 2 \sqrt{6}} \times \frac{5 - 2 \sqrt{6}}{5 - 2 \sqrt{6}} \Rightarrow \frac{1}{a} = \frac{5 - 2 \sqrt{6}}{{(5)}^{2} - {(2 \sqrt{6})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow \frac{1}{a} = \frac{5 - 2 \sqrt{6}}{25 - 24} \Rightarrow \frac{1}{a} = \frac{5 - 2 \sqrt{6}}{1} \Rightarrow \frac{1}{a} = 5 - 2 \sqrt{6} Thus, a + \frac{1}{a} = (5 + 2 \sqrt{6}) + (5 - 2 \sqrt{6}) = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} = 10$

Hence, if

$a = 5 + 2 \sqrt{6}, then a + \frac{1}{a} = 10 .$

Question 6:

$x = \sqrt{6} + \sqrt{5}, then x^{2} + \frac{1}{x^{2}} - 2 =_________.$

Answer 6:

$Given : x = \sqrt{6} + \sqrt{5} Now, \frac{1}{x} = \frac{1}{\sqrt{6} + \sqrt{5}} Multiply and divide RHS by (\sqrt{6} - \sqrt{5}), we get \Rightarrow \frac{1}{x} = \frac{1}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} \Rightarrow \frac{1}{x} = \frac{\sqrt{6} - \sqrt{5}}{{(\sqrt{6})}^{2} - {(\sqrt{5})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow \frac{1}{x} = \frac{\sqrt{6} - \sqrt{5}}{6 - 5} \Rightarrow \frac{1}{x} = \frac{\sqrt{6} - \sqrt{5}}{1} \Rightarrow \frac{1}{x} = \sqrt{6} - \sqrt{5} Thus, x^{2} + \frac{1}{x^{2}} - 2 = {(x - \frac{1}{x})}^{2} = {((\sqrt{6} + \sqrt{5}) - (\sqrt{6} - \sqrt{5}))}^{2} = {(\sqrt{6} + \sqrt{5} - \sqrt{6} + \sqrt{5})}^{2} = {(2 \sqrt{5})}^{2} = 20$

Hence, if

$x = \sqrt{6} + \sqrt{5}, then x^{2} + \frac{1}{x^{2}} - 2 = 20 .$

Question 7:

$x = 3 - \sqrt{8}, then {(x - \frac{1}{x})}^{2} =___________.$

Answer 7:

$Given : x = 3 - \sqrt{8} Now, \frac{1}{x} = \frac{1}{3 - \sqrt{8}} Multiply and divide RHS by (3 + \sqrt{8}), we get \Rightarrow \frac{1}{x} = \frac{1}{3 - \sqrt{8}} \times \frac{3 + \sqrt{8}}{3 + \sqrt{8}} \Rightarrow \frac{1}{x} = \frac{3 + \sqrt{8}}{{(3)}^{2} - {(\sqrt{8})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow \frac{1}{x} = \frac{3 + \sqrt{8}}{9 - 8} \Rightarrow \frac{1}{x} = \frac{3 + \sqrt{8}}{1} \Rightarrow \frac{1}{x} = 3 + \sqrt{8} Thus, {(x - \frac{1}{x})}^{2} = {((3 - \sqrt{8}) - (3 + \sqrt{8}))}^{2} = {(3 - \sqrt{8} - 3 - \sqrt{8})}^{2} = {(- 2 \sqrt{8})}^{2} = 32$

Hence, if

$x = 3 - \sqrt{8}, then {(x - \frac{1}{x})}^{2} = 32 .$

Question 8:

$x = \frac{2}{\sqrt{3} - \sqrt{5}} and y = \frac{2}{\sqrt{3} + \sqrt{5}}$ , then x + y = __________.

Answer 8:

$Given : x = \frac{2}{\sqrt{3} - \sqrt{5}} y = \frac{2}{\sqrt{3} + \sqrt{5}} Now, x = \frac{2}{\sqrt{3} - \sqrt{5}} Multiply and divide RHS by (\sqrt{3} + \sqrt{5}), we get \Rightarrow x = \frac{2}{\sqrt{3} - \sqrt{5}} \times \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} \Rightarrow x = \frac{2 (\sqrt{3} + \sqrt{5})}{{(\sqrt{3})}^{2} - {(\sqrt{5})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow x = \frac{2 (\sqrt{3} + \sqrt{5})}{3 - 5} \Rightarrow x = \frac{2 (\sqrt{3} + \sqrt{5})}{- 2} \Rightarrow x = \frac{- (\sqrt{3} + \sqrt{5})}{1} \Rightarrow x = - \sqrt{3} - \sqrt{5} y = \frac{2}{\sqrt{3} + \sqrt{5}} Multiply and divide RHS by (\sqrt{3} - \sqrt{5}), we get \Rightarrow y = \frac{2}{\sqrt{3} + \sqrt{5}} \times \frac{\sqrt{3} - \sqrt{5}}{\sqrt{3} - \sqrt{5}} \Rightarrow y = \frac{2 (\sqrt{3} - \sqrt{5})}{{(\sqrt{3})}^{2} - {(\sqrt{5})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow y = \frac{2 (\sqrt{3} - \sqrt{5})}{3 - 5} \Rightarrow y = \frac{2 (\sqrt{3} - \sqrt{5})}{- 2} \Rightarrow y = \frac{- (\sqrt{3} - \sqrt{5})}{1} \Rightarrow y = - \sqrt{3} + \sqrt{5} Thus, x + y = (- \sqrt{3} - \sqrt{5}) + (- \sqrt{3} + \sqrt{5}) = (- \sqrt{3} - \sqrt{5} - \sqrt{3} + \sqrt{5}) = (- 2 \sqrt{3}) = - 2 \sqrt{3}$

Hence, x + y =

$- 2 \sqrt{3}$ .

Question 9:

$\sqrt{5 + 2 \sqrt{6}} = A + \sqrt{3}, then A =_________.$

Answer 9:

$Given : \sqrt{5 + 2 \sqrt{6}} = A + \sqrt{3} Now, \sqrt{5 + 2 \sqrt{6}} = A + \sqrt{3} \Rightarrow \sqrt{2 + 3 + 2 \sqrt{6}} = A + \sqrt{3} \Rightarrow \sqrt{{(\sqrt{2})}^{2} + {(\sqrt{3})}^{2} + 2 \sqrt{6}} = A + \sqrt{3} \Rightarrow \sqrt{{(\sqrt{2} + \sqrt{3})}^{2}} = A + \sqrt{3} (Using the identity : {(a + b)}^{2} = a^{2} + b^{2} + 2 a b) \Rightarrow \sqrt{2} + \sqrt{3} = A + \sqrt{3} \Rightarrow A = \sqrt{2}$

Hence, if

$\sqrt{5 + 2 \sqrt{6}} = A + \sqrt{3}, then A = \sqrt{2} .$

Question 10:

$\sqrt{7 + 2 \sqrt{6}} - \sqrt{7 - 2 \sqrt{6}} =__________.$

Answer 10:

$\sqrt{7 + 2 \sqrt{6}} = \sqrt{6 + 1 + 2 \sqrt{6}} = \sqrt{{(\sqrt{6})}^{2} + {(1)}^{2} + 2 \sqrt{6}} = \sqrt{{(\sqrt{6} + 1)}^{2}} (Using the identity : {(a + b)}^{2} = a^{2} + b^{2} + 2 a b) = \sqrt{6} + 1 . . . . (1) \sqrt{7 - 2 \sqrt{6}} = \sqrt{6 + 1 - 2 \sqrt{6}} = \sqrt{{(\sqrt{6})}^{2} + {(1)}^{2} - 2 \sqrt{6}} = \sqrt{{(\sqrt{6} - 1)}^{2}} (Using the identity : {(a - b)}^{2} = a^{2} + b^{2} - 2 a b) = \sqrt{6} - 1 . . . . (2) From (1) and (2) \sqrt{7 + 2 \sqrt{6}} - \sqrt{7 - 2 \sqrt{6}} = (\sqrt{6} + 1) - (\sqrt{6} - 1) = \sqrt{6} + 1 - \sqrt{6} + 1 = 2$

Hence,

$\sqrt{7 + 2 \sqrt{6}} - \sqrt{7 - 2 \sqrt{6}} = 2 .$

Question 11:

$x = \frac{2}{\sqrt{10} - \sqrt{8}} and y = \frac{2}{\sqrt{10} + 2 \sqrt{2}}, then {(x - y)}^{2} =____________.$

Answer 11:

$Given : x = \frac{2}{\sqrt{10} - \sqrt{8}} y = \frac{2}{\sqrt{10} + 2 \sqrt{2}} Now, x = \frac{2}{\sqrt{10} - \sqrt{8}} Multiply and divide RHS by (\sqrt{10} + \sqrt{8}), we get \Rightarrow x = \frac{2}{\sqrt{10} - \sqrt{8}} \times \frac{\sqrt{10} + \sqrt{8}}{\sqrt{10} + \sqrt{8}} \Rightarrow x = \frac{2 (\sqrt{10} + \sqrt{8})}{{(\sqrt{10})}^{2} - {(\sqrt{8})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow x = \frac{2 (\sqrt{10} + \sqrt{8})}{10 - 8} \Rightarrow x = \frac{2 (\sqrt{10} + \sqrt{8})}{2} \Rightarrow x = \frac{(\sqrt{10} + \sqrt{8})}{1} \Rightarrow x = \sqrt{10} + \sqrt{8} \Rightarrow x = \sqrt{10} + 2 \sqrt{2} y = \frac{2}{\sqrt{10} + 2 \sqrt{2}} Multiply and divide RHS by (\sqrt{10} - 2 \sqrt{2}), we get \Rightarrow y = \frac{2}{\sqrt{10} + 2 \sqrt{2}} \times \frac{\sqrt{10} - 2 \sqrt{2}}{\sqrt{10} - 2 \sqrt{2}} \Rightarrow y = \frac{2 (\sqrt{10} - 2 \sqrt{2})}{{(\sqrt{10})}^{2} - {(2 \sqrt{2})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow y = \frac{2 (\sqrt{10} - 2 \sqrt{2})}{10 - 8} \Rightarrow y = \frac{2 (\sqrt{10} - 2 \sqrt{2})}{2} \Rightarrow y = \frac{(\sqrt{10} - 2 \sqrt{2})}{1} \Rightarrow y = \sqrt{10} - 2 \sqrt{2} Thus, {(x - y)}^{2} = {((\sqrt{10} + 2 \sqrt{2}) - (\sqrt{10} - 2 \sqrt{2}))}^{2} = {(\sqrt{10} + 2 \sqrt{2} - \sqrt{10} + 2 \sqrt{2})}^{2} = {(4 \sqrt{2})}^{2} = 32$

Hence, if

$x = \frac{2}{\sqrt{10} - \sqrt{8}} and y = \frac{2}{\sqrt{10} + 2 \sqrt{2}}, then {(x - y)}^{2} = 32 .$

Question 12:

$\sqrt{13 - x \sqrt{10}} = \sqrt{8} + \sqrt{5}, then x =__________.$

Answer 12:

$Given : \sqrt{13 - x \sqrt{10}} = \sqrt{8} + \sqrt{5} Now, \sqrt{13 - x \sqrt{10}} = \sqrt{8} + \sqrt{5} \Rightarrow \sqrt{13 - x \sqrt{10}} = \sqrt{{(\sqrt{8} + \sqrt{5})}^{2}} \Rightarrow \sqrt{13 - x \sqrt{10}} = \sqrt{{(\sqrt{8})}^{2} + {(\sqrt{5})}^{2} + 2 (\sqrt{8}) (\sqrt{5})} (Using the identity : {(a + b)}^{2} = a^{2} + b^{2} + 2 a b) \Rightarrow \sqrt{13 - x \sqrt{10}} = \sqrt{8 + 5 + 2 (2 \sqrt{2}) (\sqrt{5})} \Rightarrow \sqrt{13 - x \sqrt{10}} = \sqrt{13 + 4 \sqrt{10}} \Rightarrow - x = 4 \Rightarrow x = - 4$

Hence, if

$\sqrt{13 - x \sqrt{10}} = \sqrt{8} + \sqrt{5}, then x = - 4 .$

RD Sharma 2020 solution class 9 chapter 3 Rationalisation FBQS