Question

if $x = 3 + \sqrt{8}$then the value of$( {x}^{2} + 1 \div {x}^{2} )$is​

Answer 1

Given:-

$\sf\to x = 3+\sqrt{8}$

To find:-

$\sf\to x^2+\dfrac{1}{x^2}$

Now we can write as

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

Now rationalise

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

we get

$\sf\to\dfrac{1}{x} =3-\sqrt{8}$

Using this identities

$\to\sf (a+b)^2= a^2+b^2+2ab$

Now we can write as

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

Now put the value of x and 1/x

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

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

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

$\sf\to 36=x^2+\dfrac{1}{x^2} +2$

$\sf\to x^2+\dfrac{1}{x^2} =36-2$

$\sf\to x^2+\dfrac{1}{x^2} =34$

Answer:-

$\sf\to x^2+\dfrac{1}{x^2}$

Answer 2

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}→(3+

8

+3−

8

)

2

=x

2

+

x

2

1

\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}→(6)

2

=x

2

+

x

2

1

\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}→x

2

+

x

2

1