Question

hi !


x = 7 + 4root3

fing

rootx + 1 divided by root x

thanks

Answer 1

Hi there !

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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 = (2)² + (√3)² + 2 × 2 × √3

Using Identity :

a² + b² + 2ab = (a + b)²

⇒( 2 + √3 )²

Now,

√x = √ (2 + √3)²

√x = 2 + √3

And,

⇒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 !

Answer 2

Here is your solution

Given:-

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

We have to find :-

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

Now

Solution:-

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

Now

$\frac{1}{ \sqrt{x} } = \frac{1}{2 + \sqrt{3} } \\ now \: rationalise \\ \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{2 - \sqrt{3} }{(2 ){}^{2} - ( \sqrt{3} ) {}^{2} } \\ \\ \\ \frac{2 - \sqrt{3} }{4 - 3} \\ \\ 2 - \sqrt{3}$

$\sqrt{x} + \frac{1}{ \sqrt{x} } = 2 + \sqrt{ \cancel3} + 2 - \sqrt{\cancel3} \\ \sqrt{x} + \frac{1}{ \sqrt{x} } = 4$

Hope it helps you