Question

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$if x = 7 + 4 \sqrt{3} \: find \: the \: value \: of \: \sqrt{x} + \frac{1}{ \sqrt{x} }$

Answer 1

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Refer to the attachment above for your answer!!

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Answer 2

Hey there!

Given : x = 7 + 4√3

To find : √x + 1 / √x

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$<b><u>Solution:</b></u>$

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

We have,

x = (2 + √3)²

So,

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

And,

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

Now, coming to the question;

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


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Thanks for the question!