Question

if x=7+4√3 find the value of √x+1/√x

Answer 1

Hey mate !

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Given :


x = 7 + 4√3


To find :


√x + 1 / √x


Solution :


x = 7 + 4√3


⇒ x = 4 + 3 + 4√3


⇒ x = 2² + √3² + 2 × 2 × √3


⇒ x = ( 2 + √3)²


⇒ √x = 2 + √3


Now,


1 / √x = 1 / 2 + √3 × 2 - √3 / 2 - √3


⇒ 1 / √x = 2 - √3 / 2² - √3²


⇒ 1 / √x = 2 - √3 / 4 - 3


⇒ 1 / √x = 2 - √3


Again,


√x + 1 / √x = 2 + √3 + 2 - √3


⇒ √x + 1 / √x = 2 + 2


⇒ √x + 1 / √x = 4


Hence,


√x + 1 / √x = 4
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Thanks for the question!

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Answer 2

$\underline{\bold{Given:-}}$

x = 7 + 4 ✓3

$\underline{\bold{To\: find:-}}$

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

$\underline{\bold{Solution:-}}$

$x = 7 + 4 \sqrt{3} \\ \\ x = 4 + 3 + 2 \times 2 \times \sqrt{3} \\ \\ x = {2}^{2} + {( \sqrt{3}) }^{2} + 2 \times 2 \times \sqrt{3} \\ \\ x = {(2 + \sqrt{3} )}^{2}$

$\sqrt{x} = 2 + \sqrt{3}$

$\frac{1}{ \sqrt{x} } = \frac{1}{2 + \sqrt{3} } \\ \\ \bold{On \: rationalising \: them}\\ \\ \frac{1}{ \sqrt{x} } = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \bold{using \: identity }\\ \\ \bold{\boxed{{a}^{2} - {b}^{2} = (a + b)(a - b)}} \\ \\ \frac{1}{ \sqrt{x} } = \frac{2 - \sqrt{3} }{ {2}^{2} - { (\sqrt{3} )}^{2} } \\ \\ \frac{1}{ \sqrt{x} } = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \frac{1}{ \sqrt{x} } = 2 - \sqrt{3}$

According to question

$\sqrt{x} + \frac{1}{ \sqrt{x} } = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\\boxed{\bold{ \sqrt{x} + \frac{1}{ \sqrt{x} } = 4}}$

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