Question

If x=5-2 sqrt 6 , then find the value of sqrt(x) + 1/(sqrt(x))​

Answer 1

Answer:

10

Step-by-step explanation:

x=5-2√6

1/x=5+2√6

So, x+(1/x)=5-2√6+5+2√6=10

Answer 2

Given:

The value of x -

• $\bf \leadsto x = 5-2\sqrt{6}$

What To Find:

We have to find the value of -

• $\bf \leadsto \sqrt{x} + \dfrac{1}{\sqrt{x}}$

How To Find:

To find, we have to -

• First, find the value of $\bf \sqrt{x}$.
• Next, find the value of $\bf \dfrac{1}{\sqrt{x}}$.
• Finally, find the value of $\bf \sqrt{x} + \dfrac{1}{\sqrt{x}}$.

Solution:

• Finding the value of -

$\bf \leadsto \sqrt{x}$

We know that -

$\bf \leadsto x = 5-2\sqrt{6}$

Substitute in the expression,

$\bf \leadsto \sqrt{x} = \sqrt{5-2\sqrt{6}}$

We know that -

• $\bf \leadsto 5 = 2 + 3$
• $\bf \leadsto 6 = 2 \times 3$

Can be written as,

$\bf \leadsto \sqrt{x} = \sqrt{3+2-2 \times \sqrt{3} \times \sqrt{2}}$

Let's take $\bf 2$ as $\bf (\sqrt{2})^2$ and $\bf 3$ as $\bf (\sqrt{3})^2$,

$\bf \leadsto \sqrt{x} = \sqrt{(\sqrt{3})^2+(\sqrt{2})^2-2 \times \sqrt{3} \times \sqrt{2}}$

This is forming the identity -

$\bf \leadsto a^2 + b^2 - 2ab = (a - b)^2$

Where -

• $\bf \leadsto a = \sqrt{3}$
• $\bf \leadsto b = \sqrt{2}$

By using the identity,

$\bf \leadsto \sqrt{x} = \sqrt{(\sqrt{3}-\sqrt{2})^2}$

Simplify the RHS,

$\bf \leadsto \sqrt{x} = \sqrt{3}-\sqrt{2}$

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• Finding the value of -

$\bf \leadsto \dfrac{1}{\sqrt{x}}$

We know that -

$\bf \leadsto \sqrt{x} = \sqrt{3}-\sqrt{2}$

Substitute in the expression,

$\bf \leadsto \dfrac{1}{\sqrt{x}} = \dfrac{1}{\sqrt{3} - \sqrt{2}}$

Rationalise the denominator,

$\bf \leadsto \dfrac{1}{\sqrt{x}} = \dfrac{\sqrt{3} + \sqrt{2}}{(\sqrt{3} - \sqrt{2}) \times (\sqrt{3}+\sqrt{2})}$

Use the identity,

$\bf \leadsto (a + b) (a - b) = a^2 - b^2$

Where -

• $\bf \leadsto a = \sqrt{3}$
• $\bf \leadsto b = \sqrt{2}$

By using the identity,

$\bf \leadsto \dfrac{1}{\sqrt{x}} = \dfrac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}$

Find the squares,

$\bf \leadsto \dfrac{1}{\sqrt{x}} = \dfrac{\sqrt{3} + \sqrt{2}}{3-2}$

Subtract the values,

$\bf \leadsto \dfrac{1}{\sqrt{x}} = \dfrac{\sqrt{3} + \sqrt{2}}{1}$

Can be written as,

$\bf \leadsto \dfrac{1}{\sqrt{x}} = \sqrt{3} + \sqrt{2}$

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• Finding the value of -

$\bf \leadsto \sqrt{x} + \dfrac{1}{\sqrt{x}}$

We know that -

$\bf \leadsto \sqrt{x} = \sqrt{3}-\sqrt{2}$

$\bf \leadsto \dfrac{1}{\sqrt{x}} = \sqrt{3} + \sqrt{2}$

Substitute in the expression,

$\bf \leadsto \sqrt{x} + \dfrac{1}{\sqrt{x}} = \sqrt{3} - \sqrt{2} + \sqrt{3} + \sqrt{2}$

Rearrange the terms in RHS,

$\bf \leadsto \sqrt{x} + \dfrac{1}{\sqrt{x}} = \sqrt{3} + \sqrt{3} + \sqrt{2} - \sqrt{2}$

Solve the terms in RHS,

$\bf \leadsto \sqrt{x} + \dfrac{1}{\sqrt{x}} = 2\sqrt{3}$

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Final Answer:

∴ Thus, the value of  $\bf \sqrt{x} + \dfrac{1}{\sqrt{x}}$  is  $\bf 2\sqrt{3}$.