Question

x=1/x-5,x≠5. find x²-1/x²

Answer 1

$x = \frac{1}{x - 5} \\ {x}^{2} = \frac{1}{ {x}^{2} - 10x + 25 } \\ {x }^{2} - \frac{1}{ {x}^{2} } = \frac{1}{ {x}^{2} - 10x + 25} - ( {x}^{2} - 10x + 25) \\ = \frac{1 - ( { {x}^{2} - 10x + 25) }^{2} }{ {x}^{2} - 10x + 25} \\ = \frac{1 - ( { {(x - 5)}^{2} )}^{2} }{ {(x - 5)}^{2} }$

Answer 2

Answer :-

x² - 1/x² = 5√21

Explanation :-

Given :-

$\sf x = \dfrac{1}{5 - x}$

To find :-

x² - 1/x²

Solution :-

First find x + 1/x

$\sf x = \dfrac{1}{5 - x}$

Reciprocal on both sides

$\sf \dfrac{1}{x} = \dfrac{5 - x}{1}$

$\sf \dfrac{1}{x} = 5 - x$

Transpose - x to RHS [ - x becomes + x]

$\sf \dfrac{1}{x} + x = 5$

$\sf x + \dfrac{1}{x} = 5....(1)$

Now find x - 1/x

We know that

(x + 1/x)² - (x - 1/x)² = 4

Here x + 1/x = 5

By substituting the values

⇒ (5)² - (x - 1/x)² = 4

⇒ 25 - (x - 1/x)² = 4

⇒ - (x - 1/x)² = 4 - 25

⇒ - (x - 1/x)² = - 21

⇒ (x - 1/x)² = - (-21)

⇒ (x - 1/x)² = 21

⇒ x - 1/x = √21....(2)

From (1) and (2)

(x + 1/x)(x - 1/x) = 5(√21)

⇒ (x + 1/x)(x - 1/x) = 5√21

We know that

(a + b)(a - b) = a² - b²

Here a = x, b = 1/x

By substituting the values

⇒ (x)² - (1/x)² = 5√21

⇒ x² - 1²/x² = 5√21

⇒ x² - 1/x² = 5√21

Therefore the value of x² - 1/x² is 5√21.

Identities used :-

• (x + 1/x)² - (x - 1/x)² = 4

• (a + b)(a - b) = a² - b²