Question

solve these three questions ...



only revelant ans . needed​

Answer 1

question

$\bf \: simplify : \sqrt{3 + 2 \sqrt{2} }.$

solution

$\longmapsto \tt \sqrt{3 + 2 \sqrt{2} } \\ \\ \\\longmapsto \tt \sqrt{1 + 2 \sqrt{2} + 2}$

$\longmapsto \tt \sqrt{ {1}^{2} +( 2 \times 1 \times \sqrt{2}) + {( \sqrt{2} )}^{2} } \qquad \{ {a}^{2} + {b}^{2} - 2ab = (a + b {)}^{2} \}$

$\longmapsto \tt \cancel {\sqrt{(1 + { \sqrt{2}) }^{ \cancel2} } } \\ \\ \\ \longmapsto \tt \red{1 + \sqrt{2} } \: \: \: \: answer$

____________________

question

$\bf \: simplify: \sqrt{3 - 2 \sqrt{2} } .$

solution

$\longmapsto \tt \sqrt{3 - 2 \sqrt{2} } \\ \\ \\ \longmapsto \tt \sqrt{3 - 2 \sqrt{2 \times 1} }$

$\longmapsto \tt \sqrt{ {( \sqrt{2} }^{2}) + {(1)}^{2} - 2( \sqrt{2})(1) } \: \: \: \{ {a}^{2} + {b}^{2} - 2ab = {(a + b)}^{2} \}$

$\longmapsto \tt \cancel{ \sqrt{( \sqrt{2} - 1 {)}^{ \cancel2} } } \\ \\ \\ \longmapsto \tt \red{ \sqrt{2} - 1} \: \: \: \: \: answer$

___________________

question

$\bf \: \: if \: x = 3 + 2 \sqrt{2} \: \: then \: find \: value \: of \: \sqrt{x} - \frac{1}{ \sqrt{x} } .$

solution

given that

• $x = 3 + 2 \sqrt{2}$

to find

• $\tt \: the \: value \: of \: \sqrt{x} - \dfrac{ 1{} }{ \sqrt{x} } \: \:$

SOLUTION

$\longmapsto \tt \: x = 3 + 2 \sqrt{2} \\ \\ \\ \longmapsto \tt x = 1 + 2 \sqrt{2} + 2 \\ \\ \\ \longmapsto \tt x = {1}^{2} + 2 \times 1 \times \sqrt{2} + {( \sqrt{2}) }^{2} \\ \\ \\\longmapsto \tt x = {(1 + \sqrt{2}) }^{2} \\ \\ \\ \longmapsto \tt \sqrt{x} = 1 + \sqrt{2}$

hence the value of $\tt \red {\: { \sqrt{x} = 1 + \sqrt{2} }}$

now

$\qquad \qquad \bf \: \sqrt{x} - \dfrac{1}{ \sqrt{x} } \\ \\ \\ \\ putting \: value \: of \: \sqrt{x} = 1 + \sqrt{2}$

we get

$\qquad \qquad \:\longmapsto \tt (1 + \sqrt{2} ) - \dfrac{1}{(1 + \sqrt{2}) } \\ \\ \\ \qquad \qquad \:\longmapsto \tt \frac{ {(1 + \sqrt{2}) }^{2} - 1}{(1 + \sqrt{2} )} \\ \\ \\ \qquad \qquad \:\longmapsto \tt \frac{1 + 2\sqrt{2} + 2 - 1 }{1 + \sqrt{2} } \\ \\ \\ \qquad \qquad \:\longmapsto \tt \frac{2 \cancel{(1 + \sqrt{2} )}}{ \cancel{(1 + \sqrt{2} )}} \\ \\ \\ \qquad \qquad \:\longmapsto \tt 2$

$\large \tt\:therefore \: the \: value \: of \red{ \sqrt{x} - \dfrac{1}{ \sqrt{x} } = 2}$

Answer 2

$\tt\ \ \: 1. \: simplify \: \tt \sqrt{3 + 2 \sqrt{2} }$

$\leadsto\sf\ \sqrt{3 + 2 \sqrt{2} } \\ \\ \ \leadsto\sf\ \sqrt{1 + 2 \sqrt{2 + 2} \ } \\ \\ \leadsto\sf\ \sqrt{ {1}^{2} + (2 \times 1 \times \sqrt{2}) + \sqrt{2 {}^{2} } } \\ \\ \tt \: {a}^{2} + {b}^{2} - 2ab = (a + {b}^{2} ) \\ \\ \begin{gathered}\leadsto \sf \cancel {\sqrt{(1 + { \sqrt{2}) }^{ \cancel2} } } \\ \\ \leadsto \sf \purple{1 + \sqrt{2} } \: \: \: \:is \: the \: answer\end{gathered}$

                    ____________      

$\tt\ \ \: 2. \: simplify \: \tt \sqrt{3 - 2 \sqrt{2} }$

$\leadsto\sf\ \sqrt{3 - 2 \sqrt{2} } \\ \\ \ \leadsto\sf\ \sqrt{3 - 2 \sqrt{2 \times 1} \ } \\ \\ \leadsto\sf\ \sqrt{ {( \sqrt{2) {}^{2} + ( {1}^{2}) - 2( \sqrt{2} )(1) } } } \\ \\ \tt \: {a}^{2} + {b}^{2} - 2ab = (a + {b}^{2} ) \\ \\\begin{gathered}\leadsto \sf \cancel {\sqrt{( { \sqrt{2} - 1) }^{ \cancel2} } } \\ \\ \leadsto \sf \purple{ \sqrt{2} - 1 } \: \: \: \:is \: the \: answer\end{gathered}$

                    ____________      

$\tt\ \ \: 3. \: if \: x = 3 + 2 \sqrt{2} \: then \: find \: value \: of \: \tt \ \: \sqrt{x} - \frac{1}{ \sqrt{x} }$

$\leadsto\sf\ x = 3 + 2 \sqrt{2} \\ \\ \ \leadsto\sf\ x = 1 + 2 \sqrt{2} + 2 \\ \\ \leadsto\sf\ \ x = {1}^{2} + 2 \times 1 \times \sqrt{2} + ( \sqrt{2}) {}^{2} \\ \\ \leadsto\sf \ \ x = (1 + \sqrt{2}) {}^{2} \\ \ \\ \leadsto \sf \ \sqrt{x} - \frac{1}{ \sqrt{x} } \ \\ \\ \: \sf \ so\: put \: value \: of \: \sqrt{x} = 1 + \sqrt{2} \\ \\ \leadsto \sf (1 + \sqrt{2} ) - \frac{1}{(1 + \sqrt{2} )} \ \ \\ \ \\ \: \leadsto\sf \: \frac{(1 + \sqrt{2}) {}^{2} - 1 }{(1 + \sqrt{2}) } \\ \\ \leadsto \sf \ \frac{1 + 2 \sqrt{2} + 2 - 1 }{(1 + \sqrt{2}) }$

$\begin{gathered}\leadsto \sf \cancel { \frac{2(1 + \sqrt{2}) }{(1 + \sqrt{2}) } } \\ \\ \leadsto \sf\purple{2 } \: \: \: \: \\ \\ \sf\ therefore \: value \:of \: \sqrt{x} - \frac{1}{ \sqrt{x} } = 2 \end{gathered}$

                    ____________