Question

solve this question please​

Answer 1

Given

$\to \sf \: x - \dfrac{1}{x} = \dfrac{1}{2}$

To Find the Value

$\sf \to \: 4 {x}^{2} + \dfrac{4}{ {x}^{2} }$

Now Take

$\to \sf \: x - \dfrac{1}{x} = \dfrac{1}{2}$

Squaring on both side

$\to \sf \: \bigg( x - \dfrac{1}{x} \bigg) ^{2} = \bigg(\dfrac{1}{2} \bigg)^{2}$

Use This identities

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

we get

$\sf \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times x \times \dfrac{1}{x} = \dfrac{1}{4}$

$\sf \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = \dfrac{1}{4}$

$\sf \to \: 4 \bigg({x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \bigg) = 1$

$\sf \to \: 4{x}^{2} + \dfrac{4}{ {x}^{2} } - 8 = 1$

$\sf \to \: 4{x}^{2} + \dfrac{4}{ {x}^{2} } = 1 + 8$

$\sf \to \: 4{x}^{2} + \dfrac{4}{ {x}^{2} } = 9$

Answer

$\sf \to \: 4{x}^{2} + \dfrac{4}{ {x}^{2} } = 9$