Question

X+1/X=6 find 1.x-1/x​

Answer 1

Given :-

$\rm :\longmapsto\:x + \dfrac{1}{x} = 6$

To Find :-

$\rm :\longmapsto\:x - \dfrac{1}{x}$

Solution :-

Given that

$\rm :\longmapsto\:x + \dfrac{1}{x} = 6$

We know,

$\red{\boxed{ \bf\: {(a + b)}^{2} - {(a - b)}^{2} = 4ab}}$

In this identity,

$\rm :\longmapsto\:Replace \: a \: by \: x \: \: \: and \: \: \: b \: by \:\dfrac{1}{x}$

So, we get

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

$\rm :\longmapsto\: {(6)}^{2} - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 4$

$\rm :\longmapsto\: 36 - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 4$

$\rm :\longmapsto\: - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 4 - 36$

$\rm :\longmapsto\: - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = - 32$

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 32$

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} \bigg) }= \pm \: \sqrt{32}$

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

Hence,

$\: \: \: \: \: \: \: \: \: \: \boxed{\bf :\implies\: {\bigg(x - \dfrac{1}{x} \bigg) } = \pm \: 4 \sqrt{2}}$

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)