Question

$\displaystyle\sf x = 3+2\sqrt{2}$
$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$??
​

Answer 1

Step-by-step explanation:

Given

\sf\to x = 3+\sqrt{8}→x=3+

8

To find

\sf\to x^2+\dfrac{1}{x^2}→x

2

+

x

2

1

Now we can write as

\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}→

x

1

=

3+

8

1

Now rationalise

\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}\times\dfrac{3-\sqrt{8}}{3-\sqrt{8}}=\dfrac{3-\sqrt{8}}{9-8} =3-\sqrt{8}→

x

1

=

3+

8

1

×

3−

8

3−

8

=

9−8

3−

8

=3−

8

we get

\sf\to\dfrac{1}{x} =3-\sqrt{8}→

x

1

=3−

8

Using this identities

\to\sf (a+b)^2= a^2+b^2+2ab→(a+b)

2

=a

2

+b

2

+2ab

Now we can write as

\sf\to\bigg(x+\dfrac{1}{x}\bigg)^2=x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}→(x+

x

1

)

2

=x

2

+

x

2

1

+2×x×

x

1

Now put the value of x and 1/x

\sf\to (3+\sqrt{8}+3-\sqrt{8})^2=x^2+\dfrac{1}{x^2} +2→(3+

8

+3−

8

)

2

=x

2

+

x

2

1

+2

\sf\to (3+3)^2=x^2+\dfrac{1}{x^2} +2→(3+3)

2

=x

2

+

x

2

1

+2

\sf\to (6)^2=x^2+\dfrac{1}{x^2} +2→(6)

2

=x

2

+

x

2

1

+2

\sf\to 36=x^2+\dfrac{1}{x^2} +2→36=x

2

+

x

2

1

+2

\sf\to x^2+\dfrac{1}{x^2} =36-2→x

2

+

x

2

1

=36−2

\sf\to x^2+\dfrac{1}{x^2} =34→x

2

+

x

2

1

=34

Answer

\sf\to x^2+\dfrac{1}{x^2} =34→x

2

+

x

2

1

=34

Answer 2

Step-by-step explanation:

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

$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$

•••••••••••••••••••••••••••••••••••••••••••••••••

$\begin{gathered}\displaystyle\sf :\implies \dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}}\\\end{gathered}$

$\displaystyle\sf :\implies \dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{3^2-(2\sqrt{2}^2)}$

$\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{9-8}$

$\displaystyle\sf :\implies \dfrac{1}{x} = 3-2\sqrt{2}$

•••••••••••••••••••••••••••••••••••••••••••••••••

$\displaystyle\sf :\implies x+\dfrac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2})$

$\displaystyle\sf :\implies x+\dfrac{1}{x} = 3+2\sqrt{2} + 3 - 2\sqrt{2}$

$\displaystyle\sf :\implies x+\dfrac{1}{x}$

So here we know that we may split the number 6 into 4+2 and 4+2 = 6

$\displaystyle\sf :\implies x+\dfrac{1}{x} = 4$

$\displaystyle\sf :\implies x+\dfrac{1}{x}-2 = 4$

$\displaystyle\sf :\implies \bigg\lgroup \sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg\rgroup^2 = 4$

$\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \sqrt{4}$

$\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \pm 2$

$\displaystyle\therefore\:\underline{\textsf{The value of \sqrt{ \sf x}-\dfrac{\sf 1}{\sqrt{\sf x}} is \textbf{ \pm 2 }}}$