Question

Genius please answer on a notebook ​

Answer 1

$\sf{x = 3 - 2 \sqrt{2} }$

$\sf \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} }$

rationalise the denominator

$\sf \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} }$

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

$\sf \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{9- 8}$

$\sf \frac{1}{x} = {3 + 2 \sqrt{2} }$

now,

$\sf{x - \frac{1}{x} = 3 - 2 \sqrt{2} - (3 + 2 \sqrt{2} ) }$

$\sf{x - \frac{1}{x} = 3 - 2 \sqrt{2} - 3 - 2 \sqrt{2} }$

$\fbox{\sf{x - \frac{1}{x} = - 4 \sqrt{2} }}$

$\sf{x + \frac{1}{x} = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} }$

$\fbox{\star\:\sf{x + \frac{1}{x} = 6}}$

$\sf{ {x}^{2} - \frac{1}{ {x}^{2} } = (x + \frac{1}{x} )(x - \frac{1}{x} ) }$

$\sf{ {x}^{2} - \frac{1}{ {x}^{2} } = (6)( - 4 \sqrt{2} ) }$

$\fbox{ \sf{ {x}^{2} - \frac{1}{ {x}^{2} } = - 24 \sqrt{2}} }$

Answer 2

$\mathfrak{\large{\underline{\underline{Answer :}}}}$

$\boxed{{x}^{2} - \frac{1}{{x}^{2}} = - 24\sqrt{2}}$

$\mathfrak{\large{\underline{\underline{Explanation :}}}}$

Refer to the attachment.

$\textbf{\large{\underline{\underline{Algebraic Identities used :}}}}$

• (x + y)² = x² + 2xy + y²

• (x + y)(x - y) = x² - y²

$\textsf{\large{\underline{\underline{Additional Information :}}}}$

• An Algebraic identity is an equation, which is true for all values of the variables in the equation.

1) (x + y)² = x² + 2xy + y²

2) (x - y)² = x² - 2xy - y²

3) (x + y)(x - y) = x² - y²

4) (x + a)(x + b) = x² + (a + b)x + ab

5) (x + y)³ = x³ + 3x²y + 3xy² + y³

6) (x - y)³ = x³ - 3x²y + 3xy² - y³

7) x³ + y³ = (x + y)(x² - xy + y²)

8) x³ - y³ = (x - y)(x² + xy + y²)

9) (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz

10) x³ + y³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz

• If the product of two two irrational numbers is rational number then each of the two is the Rationalising factor of the other.