Question

rationalise these numderx+1/x=√7 ,x²+1/x²​

Answer 1

Given

$\sf \to \: x + \dfrac{1}{x} = \sqrt{7}$

To Find the value of

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

Now Take

$\sf \to \: x + \dfrac{1}{x} = \sqrt{7}$

Squaring on both side

$\sf \to \bigg(x + \dfrac{1}{x} \bigg) ^{2} = ( \sqrt{7} )^{2}$

Using this identities

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

We get

$\sf \to {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 7$

$\sf \to {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times \cancel{x}\times \dfrac{1}{ \cancel{x}} = 7$

$\sf \to {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 7$

$\sf \to {x}^{2} + \dfrac{1}{ {x}^{2} } = 7 - 2$

$\sf \to {x}^{2} + \dfrac{1}{ {x}^{2} } = 5$

Answer

$\sf \to {x}^{2} + \dfrac{1}{ {x}^{2} } = 5$