Question

x की power 2 plus 1 by x की power 2 equal to 3, then minus x minus 1/x ?​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \: \: \: \bull \: \sf \: \: \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 3$

$\large\underline{\sf{To\:Find - }}$

$\: \: \: \: \: \: \: \: \bull \: \sf \: \: \: x - \dfrac{1}{x}$

$\large\underline{\sf{Solution-}}$

We know that,

$\rm :\longmapsto\: {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$\boxed{ \: \: \: \: \bf \: Put \: a = x \: \: and \: \: b \: = \: \dfrac{1}{x} \: we \: get \: \: \: \: }$

$\rm :\longmapsto\:{\bigg(x + \dfrac{1}{x} \bigg) }^{2} - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 4 \times x \times \dfrac{1}{x}$

$\rm :\longmapsto\: {(3)}^{2} - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 4$

$\rm :\longmapsto\:9 - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 4$

$\rm :\longmapsto\:9 - 4 = {\bigg(x - \dfrac{1}{x} \bigg) }^{2}$

$\rm :\longmapsto\:5 = {\bigg(x - \dfrac{1}{x} \bigg) }^{2}$

$\bf\implies \:x - \dfrac{1}{x} = \pm \: \sqrt{5}$

Additional Information :-

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)