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:

$$\begin{gathered} \frac{1}{\sqrt{7}+2} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{by} \left(\sqrt{7}-2\right), \mathrm{we} \mathrm{get} \\ =\frac{1}{\sqrt{7}+2}\times \frac{\sqrt{7}-2}{\sqrt{7}-2} \\ =\frac{\sqrt{7}-2}{{\left(\sqrt{7}\right)}^2-2^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ =\frac{\sqrt{7}-2}{7-4} \\ =\frac{\sqrt{7}-2}{3} \end{gathered}$$

Hence, the number obtained by rationalizing the denominator of $\frac{1}{\sqrt{7}+2}$ is $\underline{ \frac{\sqrt{7}-2}{3} }.$.

Question 2:

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

Answer 2:

$$\begin{gathered} \frac{1}{\sqrt{9}-\sqrt{8}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{by} \left(\sqrt{9}+\sqrt{8}\right), \mathrm{we} \mathrm{get} \\ =\frac{1}{\sqrt{9}-\sqrt{8}}\times \frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}} \\ =\frac{\sqrt{9}+\sqrt{8}}{{\left(\sqrt{9}\right)}^2-{\left(\sqrt{8}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ =\frac{\sqrt{9}+\sqrt{8}}{9-8} \\ =\frac{\sqrt{9}+\sqrt{8}}{1} \\ =\sqrt{9}+\sqrt{8} \\ =3+2\sqrt{2} \\ \mathrm{Now}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \\ \frac{1}{\sqrt{9}-\sqrt{8}}=A+B\sqrt{2} \\ \Rightarrow 3+2\sqrt{2}=A+B\sqrt{2} \\ \Rightarrow A=3 \mathrm{and} B=2 \end{gathered}$$


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:

$$\begin{gathered} \frac{7}{3\sqrt{3}-2\sqrt{2}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{by} \left(3\sqrt{3}+2\sqrt{2}\right), \mathrm{we} \mathrm{get} \\ =\frac{7}{3\sqrt{3}-2\sqrt{2}}\times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}} \\ =\frac{7\left(3\sqrt{3}+2\sqrt{2}\right)}{{\left(3\sqrt{3}\right)}^2-{\left(2\sqrt{2}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ =\frac{21\sqrt{3}+14\sqrt{2}}{27-8} \\ =\frac{21\sqrt{3}+14\sqrt{2}}{19} \end{gathered}$$


Hence, after rationalizing the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$, we get the denominator as $\underline{\frac{21\sqrt{3}+14\sqrt{2}}{19}}.$

Question 4:

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

Answer 4:

$$\begin{gathered} \mathrm{Given}: a=2+\sqrt{3} \\ \mathrm{Now}, \frac{1}{a}=\frac{1}{2+\sqrt{3}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(2-\sqrt{3}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{1}{a}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ \Rightarrow \frac{1}{a}=\frac{2-\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{a}=\frac{2-\sqrt{3}}{4-3} \\ \Rightarrow \frac{1}{a}=\frac{2-\sqrt{3}}{1} \\ \Rightarrow \frac{1}{a}=2-\sqrt{3} \\ \mathrm{Thus}, \\ a-\frac{1}{a}=\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right) \\ =2+\sqrt{3}-2+\sqrt{3} \\ =2\sqrt{3} \end{gathered}$$


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

Question 5:

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

Answer 5:

$$\begin{gathered} \mathrm{Given}: a=5+2\sqrt{6} \\ \mathrm{Now}, \frac{1}{a}=\frac{1}{5+2\sqrt{6}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(5-2\sqrt{6}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{1}{a}=\frac{1}{5+2\sqrt{6}}\times \frac{5-2\sqrt{6}}{5-2\sqrt{6}} \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{{\left(5\right)}^2-{\left(2\sqrt{6}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{25-24} \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{1} \\ \Rightarrow \frac{1}{a}=5-2\sqrt{6} \\ \mathrm{Thus}, \\ a+\frac{1}{a}=\left(5+2\sqrt{6}\right)+\left(5-2\sqrt{6}\right) \\ =5+2\sqrt{6}+5-2\sqrt{6} \\ =10 \end{gathered}$$


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

Question 6:

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

Answer 6:

$$\begin{gathered} \mathrm{Given}: x=\sqrt{6}+\sqrt{5} \\ \mathrm{Now}, \frac{1}{x}=\frac{1}{\sqrt{6}+\sqrt{5}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{6}-\sqrt{5}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{1}{x}=\frac{1}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{5}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{1} \\ \Rightarrow \frac{1}{x}=\sqrt{6}-\sqrt{5} \\ \mathrm{Thus}, \\ {x}^2+\frac{1}{x^2}-2={\left(x-\frac{1}{x}\right)}^2 \\ ={\left(\left(\sqrt{6}+\sqrt{5}\right)-\left(\sqrt{6}-\sqrt{5}\right)\right)}^2 \\ ={\left(\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}\right)}^2 \\ ={\left(2\sqrt{5}\right)}^2 \\ =20 \end{gathered}$$


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

Question 7:

If $x=3-\sqrt{8}, \mathrm{then} {\left(x-\frac{1}{x}\right)}^2= \_\_\_\_\_\_\_\_\_\_\_.$

Answer 7:

$$\begin{gathered} \mathrm{Given}: x=3-\sqrt{8} \\ \mathrm{Now}, \frac{1}{x}=\frac{1}{3-\sqrt{8}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(3+\sqrt{8}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{1}{x}=\frac{1}{3-\sqrt{8}}\times \frac{3+\sqrt{8}}{3+\sqrt{8}} \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{{\left(3\right)}^2-{\left(\sqrt{8}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{9-8} \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{1} \\ \Rightarrow \frac{1}{x}=3+\sqrt{8} \\ \mathrm{Thus}, \\ {\left(x-\frac{1}{x}\right)}^2={\left(\left(3-\sqrt{8}\right)-\left(3+\sqrt{8}\right)\right)}^2 \\ ={\left(3-\sqrt{8}-3-\sqrt{8}\right)}^2 \\ ={\left(-2\sqrt{8}\right)}^2 \\ =32 \end{gathered}$$


Hence, if $x=3-\sqrt{8}, \mathrm{then} {\left(x-\frac{1}{x}\right)}^2=\underline{32}.$

Question 8:

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

Answer 8:

$$\begin{gathered} \mathrm{Given}: \\ x=\frac{2}{\sqrt{3}-\sqrt{5}} \\ y=\frac{2}{\sqrt{3}+\sqrt{5}} \\ \mathrm{Now}, \\ x=\frac{2}{\sqrt{3}-\sqrt{5}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{3}+\sqrt{5}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow x=\frac{2}{\sqrt{3}-\sqrt{5}}\times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}} \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{5}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{3-5} \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{-2} \\ \Rightarrow x=\frac{-\left(\sqrt{3}+\sqrt{5}\right)}{1} \\ \Rightarrow x=-\sqrt{3}-\sqrt{5} \\ y=\frac{2}{\sqrt{3}+\sqrt{5}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{3}-\sqrt{5}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow y=\frac{2}{\sqrt{3}+\sqrt{5}}\times \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{5}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{3-5} \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{-2} \\ \Rightarrow y=\frac{-\left(\sqrt{3}-\sqrt{5}\right)}{1} \\ \Rightarrow y=-\sqrt{3}+\sqrt{5} \\ \mathrm{Thus}, \\ x+y=\left(-\sqrt{3}-\sqrt{5}\right)+\left(-\sqrt{3}+\sqrt{5}\right) \\ =\left(-\sqrt{3}-\sqrt{5}-\sqrt{3}+\sqrt{5}\right) \\ =\left(-2\sqrt{3}\right) \\ =-2\sqrt{3} \end{gathered}$$


Hence, x + y = $\underline{-2\sqrt{3}}$.

Question 9:

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

Answer 9:

$$\begin{gathered} \mathrm{Given}: \\ \sqrt{5+2\sqrt{6}}=A+\sqrt{3} \\ \mathrm{Now}, \\ \sqrt{5+2\sqrt{6}}=A+\sqrt{3} \\ \Rightarrow \sqrt{2+3+2\sqrt{6}}=A+\sqrt{3} \\ \Rightarrow \sqrt{{\left(\sqrt{2}\right)}^2+{\left(\sqrt{3}\right)}^2+2\sqrt{6}}=A+\sqrt{3} \\ \Rightarrow \sqrt{{\left(\sqrt{2}+\sqrt{3}\right)}^2}=A+\sqrt{3} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}{\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow \sqrt{2}+\sqrt{3}=A+\sqrt{3} \\ \Rightarrow A=\sqrt{2} \end{gathered}$$


Hence, if $\sqrt{5+2\sqrt{6}}=A+\sqrt{3}, \mathrm{then} A=\underline{\sqrt{2}}.$

Question 10:

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

Answer 10:

$$\begin{gathered} \sqrt{7+2\sqrt{6}} \\ =\sqrt{6+1+2\sqrt{6}} \\ =\sqrt{{\left(\sqrt{6}\right)}^2+{\left(1\right)}^2+2\sqrt{6}} \\ =\sqrt{{\left(\sqrt{6}+1\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}{\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ =\sqrt{6}+1 ....\left(1\right) \\ \sqrt{7-2\sqrt{6}} \\ =\sqrt{6+1-2\sqrt{6}} \\ =\sqrt{{\left(\sqrt{6}\right)}^2+{\left(1\right)}^2-2\sqrt{6}} \\ =\sqrt{{\left(\sqrt{6}-1\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}{\left(a-b\right)}^2=a^2+b^2-2ab\right) \\ =\sqrt{6}-1 ....\left(2\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right) \\ \sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}=\left(\sqrt{6}+1\right)-\left(\sqrt{6}-1\right) \\ =\sqrt{6}+1-\sqrt{6}+1 \\ =2 \end{gathered}$$


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

Question 11:

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

Answer 11:

$$\begin{gathered} \mathrm{Given}: \\ x=\frac{2}{\sqrt{10}-\sqrt{8}} \\ y=\frac{2}{\sqrt{10}+2\sqrt{2}} \\ \mathrm{Now}, \\ x=\frac{2}{\sqrt{10}-\sqrt{8}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{10}+\sqrt{8}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow x=\frac{2}{\sqrt{10}-\sqrt{8}}\times \frac{\sqrt{10}+\sqrt{8}}{\sqrt{10}+\sqrt{8}} \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{{\left(\sqrt{10}\right)}^2-{\left(\sqrt{8}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{10-8} \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{2} \\ \Rightarrow x=\frac{\left(\sqrt{10}+\sqrt{8}\right)}{1} \\ \Rightarrow x=\sqrt{10}+\sqrt{8} \\ \Rightarrow x=\sqrt{10}+2\sqrt{2} \\ y=\frac{2}{\sqrt{10}+2\sqrt{2}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{10}-2\sqrt{2}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow y=\frac{2}{\sqrt{10}+2\sqrt{2}}\times \frac{\sqrt{10}-2\sqrt{2}}{\sqrt{10}-2\sqrt{2}} \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{{\left(\sqrt{10}\right)}^2-{\left(2\sqrt{2}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{10-8} \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{2} \\ \Rightarrow y=\frac{\left(\sqrt{10}-2\sqrt{2}\right)}{1} \\ \Rightarrow y=\sqrt{10}-2\sqrt{2} \\ \mathrm{Thus}, \\ {\left(x-y\right)}^2={\left(\left(\sqrt{10}+2\sqrt{2}\right)-\left(\sqrt{10}-2\sqrt{2}\right)\right)}^2 \\ ={\left(\sqrt{10}+2\sqrt{2}-\sqrt{10}+2\sqrt{2}\right)}^2 \\ ={\left(4\sqrt{2}\right)}^2 \\ =32 \end{gathered}$$


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

Question 12:

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

Answer 12:

$$\begin{gathered} \mathrm{Given}: \\ \sqrt{13-x\sqrt{10}}=\sqrt{8}+\sqrt{5} \\ \mathrm{Now}, \\ \sqrt{13-x\sqrt{10}}=\sqrt{8}+\sqrt{5} \\ \Rightarrow \sqrt{13-x\sqrt{10}}=\sqrt{{\left(\sqrt{8}+\sqrt{5}\right)}^2} \\ \Rightarrow \sqrt{13-x\sqrt{10}}=\sqrt{{\left(\sqrt{8}\right)}^2+{\left(\sqrt{5}\right)}^2+2\left(\sqrt{8}\right)\left(\sqrt{5}\right)} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}{\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow \sqrt{13-x\sqrt{10}}=\sqrt{8+5+2\left(2\sqrt{2}\right)\left(\sqrt{5}\right)} \\ \Rightarrow \sqrt{13-x\sqrt{10}}=\sqrt{13+4\sqrt{10}} \\ \Rightarrow -x=4 \\ \Rightarrow x=-4 \end{gathered}$$


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