Question

9. If x = 7 +4v3, find the value of (√x+1/√x).​

Answer 1

Answer:

$\boxed{\huge{\pink{ \sqrt{x} + \dfrac{1}{ \sqrt{x} } = 4}}}$

Step-by-step explanation:

$\bold{\red{Given}}\longrightarrow \\ x = 7 + 4 \sqrt{3}$

$\bold{\green{To \: find}}\longrightarrow \\ value \: of \: \\ (\sqrt{x} + \dfrac{1}{ \sqrt{x} } )$

$\bold{\blue{Concept \: used}}\longrightarrow \\ 1) {x}^{2} + {y}^{2} + 2xy = {(x + y)}^{2} \\ 2) {x}^{2} - {y}^{2} = (x + y) \: (x - y)$

$\bold{\pink{Solution}}\longrightarrow \\ x = 7 + 4 \sqrt{3} \\ \: \: \: \: = 4 + 3 + 4 \sqrt{3} \\ \: \: \: \: = {(2)}^{2} + {( \sqrt{3}) }^{2} + 2 \: (2) \: ( \sqrt{3} ) \\ x = {(2 + \sqrt{3} )}^{2} \\ taking \: square \: root \: of \: both \: sides \\ \sqrt{x} = \sqrt{ {(2 + \sqrt{3} )}^{2} } \\ = > \sqrt{x} = 2 + \sqrt{3} \\ = > \dfrac{1}{ \sqrt{x} } = \dfrac{1}{2 + \sqrt{3} } \\ = > \dfrac{1}{ \sqrt{x} } = \dfrac{1(2 + \sqrt{3}) }{(2 + \sqrt{3}) \: (2 - \sqrt{3}) } \\ = > \dfrac{1}{ \sqrt{x} } = \dfrac{(2 + \sqrt{3}) }{ {(2)}^{2} - { (\sqrt{3} )}^{2} } \\ = > \dfrac{1}{ \sqrt{x} } = \dfrac{(2 + \sqrt{3} )}{4 - 3} \\ = > \dfrac{1}{ \sqrt{x} } = \dfrac{(2 + \sqrt{3}) }{1} \\ = > \dfrac{1}{ \sqrt{x} } = 2 + \sqrt{3}$

$now$

$\sqrt{x} + \dfrac{1}{ \sqrt{x} } = (2 + \sqrt{3} ) + (2 - \sqrt{3} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4$

Answer 2

Given:

$x = 7 + 4\sqrt{3}$

To Find :

$( \sqrt{x} + \frac{1}{ \sqrt{x} } )$

Solution:

$x = 7 + 4 \sqrt{3} \\ \\ = 4 + 3 + 4 \sqrt{3} \\ \\ = ( {2)}^{2} + ( \sqrt{3)^{2} } + 2(2)( \sqrt{3} ) \\ \\ x = (2 + \sqrt{3)}^{2} \\ \\ \sqrt{x} = \sqrt{(2 + \sqrt{3)} ^{2} } \\ \\ \sqrt{x} = 2 + \sqrt{3} \\ \\ \frac{1}{ \sqrt{x} } = \frac{1}{2 + \sqrt{3} } \\ \\ \frac{1}{ \sqrt{x} } = \frac{(2 + \sqrt{3)} }{ {(2)}^{2} - ( \sqrt{3} )^{2} } \\ \\ \frac{1}{ \sqrt{x}} = \frac{(2 + \sqrt{3)} }{4 - 3} \\ \\ \frac{1}{ \sqrt{x} } = \frac{(2 + \sqrt{3} }{1} \\ \\ \frac{1}{ \sqrt{x} } = 2 + \sqrt{3} \\ \\ = 4$