Question

if x = 3 + \sqrt{8}x=3+8​ then the value of( {x}^{2} + 1 \div {x}^{2} )(x2+1÷x2) is​

Answer 1

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:-

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

Answer 2

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:-

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