English, asked by virtuoso14, 2 months ago

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

Answers

Answered by Anonymous
1

Answer:

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

Answered by abhishek917211
0

Explanation:

Answer:

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

\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} +2

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

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

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

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

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

Answer:-

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

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