Question

If x =3+2√2then find √x +1/√x

Answer 1

Answer:

2√2

Step-by-step explanation:

⇒ x = 3 + 2√2

⇒ x = 2 + 1 + 2√2

⇒ x = ( √2 )^2 + 1^2 + 2( √2 )( 1 )

⇒ x = ( √2 + 1 )^2              { using a^2 + b^2 + 2ab = ( a + b )^2 }

⇒ √x = √2 + 1

    Therefore,

      ⇒ 1 / √x = 1 / ( √2 + 1 )

  Multiplying as well as divide by √2 - 1:

      ⇒ 1 / √x = ( √2 - 1 ) / ( √2 + 1 )( √2 - 1 )

                    = ( √2 - 1 ) / ( 2 - 1 )           { ( a + b )( a - b ) = a^2 - b^2 }

                    = √2 - 1

Therefore,

⇒ √x + 1 / √x = √2 + 1 + √2 - 1

⇒ √x + 1 / √x = 2√2

Answer 2

GiVeN : -

• x = 3 + 2√2

To FiNd : -

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

SoLuTiOn : -

Here,

x = 3 + 2√2. ,

Then,

$\implies \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \\ \implies \: \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \implies \: \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2)} }^{2} } \\ \\ \implies \: \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} = 3 - 2 \sqrt{2}$

Now,

We have

=> x = 3 + 2√2 and 1/x = 3 - 2√2

Now,

We know that,

$\implies \: {( \sqrt{x} + \frac{1}{ \sqrt{x} } ) }^{2} = x + \frac{1}{x} + 2$

So,

Put the above values in the above identity

We get,

$\implies \: {( \sqrt{x} + \frac{1}{ \sqrt{x} } ) }^{2} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} + 2 \\ \\ \implies \: {( \sqrt{x} + \frac{1}{ \sqrt{x} } )}^{2} =8 \\ \\ \implies \: \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{8} = 2 \sqrt{2}$

So, we get,

=> √x + 1/√x = 2√2