Question

If x+1/x=3 then find value of x-1/x=? And x^4 - 1/x^4=?​

Answer 1

EXPLANATION.

⇒ (x + 1/x) = 3.

As we know that,

Squaring on both sides of the equation, we get.

⇒ (x + 1/x)² = (3)².

⇒ (x)² + (1/x)² + 2(x)(1/x) = 9.

⇒ x² + 1/x² + 2 = 9.

⇒ x² + 1/x² = 9 - 2.

⇒ x² + 1/x² = 7.

As we know that,

Formula of :

⇒ (x - y)² = x² + y² - 2xy.

Using this formula in the equation, we get.

⇒ (x - 1/x)² = (x)² + (1/x)² - 2(x)(1/x).

⇒ (x - 1/x)² = x² + 1/x² - 2.

Put the values of (x² + 1/x² = 7) in the equation, we get.

⇒ (x - 1/x)² = 7 - 2.

⇒ (x - 1/x)² = 5.

⇒ (x - 1/x) = √5.

⇒ (x⁴ - 1/x⁴) = (x² + 1/x²)(x² - 1/x²).

⇒ (x⁴ - 1/x⁴) = (x² + 1/x²)(x - 1/x)(x + 1/x).

Put the values in the equation, we get.

⇒ (x⁴ - 1/x⁴) = (7)(√5)(3).

⇒ (x⁴ - 1/x⁴) = 21√5.

Answer 2

$\large \underline{ \underline{ \text{Question:}}}$

• If x + 1/x = 3 then find value of x - 1/x = ? And x⁴ - 1/x⁴ = ?

$\large \underline{ \underline{ \text{Solution:}}}$

As given,

$\longrightarrow x + \frac{1}{x} = 3$

As we know that,

$[ {(x - y)}^{2} = {(x + y)}^{2} - 4xy ]$

So we can say that,

$\longrightarrow { \bigg(x - \frac{1}{x} \bigg) }^{2} = { \bigg(x + \frac{1}{x} \bigg) }^{2} - 4(x)( \frac{1}{x} )$

Let substitute the given value,

$\longrightarrow { \bigg(x - \frac{1}{x} \bigg) }^{2} = {3}^{2} - 4$

$\longrightarrow { \bigg(x - \frac{1}{x} \bigg) }^{2} = 9 - 4$

$\longrightarrow { \bigg(x - \frac{1}{x} \bigg) }^{2} = {( \sqrt{5} )}^{2}$

Comparing both sides,

$\longrightarrow \boxed{x - \frac{1}{x} = \sqrt{5}}$

As we know that,

$[ {x}^{2} - {y}^{2} = (x + y)(x - y) ]$

So we can say that,

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

Let substitute the values,

$\longrightarrow {x}^{2} - \frac{1}{ {x}^{2} } = 3 \sqrt{5} \: \: \: \: \: ...(1)$

We have,

$\longrightarrow x + \frac{1}{x} = 3$

Squaring both sides,

$\longrightarrow { \bigg(x + \frac{1}{x} \bigg)}^{2} = {3}^{2}$

As we know that,

$[ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy ]$

So we can say that,

$\longrightarrow {x}^{2} + 2(x)( \frac{1}{x} ) + { \bigg(\frac{1}{x} \bigg)}^{2} = 9$

We get,

$\longrightarrow {x}^{2} + 2+ \frac{1}{ {x}^{2} } = 9$

$\longrightarrow {x}^{2} + \frac{1}{ {x}^{2} } = 9 - 2$

$\longrightarrow {x}^{2} + \frac{1}{ {x}^{2} } = 7 \: \: \: \: \: ...(2)$

As we know that,

$[ {x}^{2} - {y}^{2} = (x + y)(x - y) ]$

So we can say that,

$\longrightarrow {x}^{4} - \frac{1}{ {x}^{4} } = \bigg({x}^{2} + \frac{1}{ {x}^{2}} \bigg ) \bigg( {x}^{2} - \frac{1}{ {x}^{2} } \bigg)$

Let substitute the values by (1) and (2),

$\longrightarrow {x}^{4} - \frac{1}{ {x}^{4} } = 7 \times (3 \sqrt{5})$

We get,

$\longrightarrow \boxed{{x}^{4} - \frac{1}{ {x}^{4} } = 21 \sqrt{5}}$