Question

if x =7-4√3 find value of √x+1/√x

Answer 1

It is given that the value of x is 7 - 4√3.


= > x = 7 - 4√3 ---: ( 1 )



So,
$\implies \dfrac{1}{x} = \dfrac{1}{7 - 4 \sqrt{3} }$



By Rationalizing, divide and multiply by ( 7 + 4√3 ) on right hand side,

$\implies \dfrac{1}{x} = \dfrac{1}{7 - 4 \sqrt{3} } \times \dfrac{7 +4 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ \\ \implies \frac{1}{x} = \frac{7 + 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3} )}^{2} } \quad \quad \mathsf{ \bigg |(a + b)(a - b) = {a}^{2} - {b}^{2} } \\ \\ \\ \implies \dfrac{1}{x} = \dfrac{7 + 4 \sqrt{3} }{49 - 48} \\ \\ \\ \implies \dfrac{1}{x} = 7 + 4 \sqrt{3} \: \: \: \: - - - \: : (2)$



Then, adding ( 1 ) and ( 2 ),

$\implies x + \dfrac{1}{x} = 7 - 4 \sqrt{3} + 7 + 4 \sqrt{3} \\ \\ \implies x + \dfrac{1}{x} = 14$



Adding 2 on both sides,

$\implies x + \dfrac{1}{x} + 2 = 14 + 2 \\ \\ \implies x + \dfrac{1}{x} + 2 \bigg( \sqrt{x} \times \dfrac{1}{ \sqrt{x} } \bigg) = 16$



We know,
( √x )² = x And similarly, ( 1 / √x )² = 1 / x


$\implies {( \sqrt{x}) }^{2} + \dfrac{1}{( \sqrt{ {x})}^{2} } + 2 \bigg( \sqrt{x} \times \dfrac{1}{ \sqrt{x} } \bigg) = 4 \times 4$


=====================
From the properties of factorization
a² + b² + 2ab = ( a + b )²
=====================


$\implies \bigg( \sqrt{x} + \dfrac{1}{ \sqrt{x} } \bigg) {}^{2} = 4 {}^{2} \\ \\ \\ \implies \sqrt{x } + \dfrac{1}{ \sqrt{x} } = 4 \: or \: - 4 \quad \quad \quad \mathsf{ \bigg |\because \sqrt{16} = \pm4}$



Hence,
the numeric value of $\sqrt{x}+\dfrac{1}{\sqrt{x}}$ is 4, taking only positive value.