Question

if (x+1/x^2)=6, find the values of
(i) (x-1/x)
(ii) (x^2 - 1/x^2)

Answer 1

Solution -

Given :

(x+1/x)² = 6

> x² + 1/x² + 2(x)(1/x) = 6

> x² + 1/x² + 2 = 6

> x² + 1/x² = 4

Subtracting 2 both sides

> x² + 1/x² - 2 = 4-2 = 2

> ( x² + 1/x² - 2(x)(1/x)) = 2

> ( x - 1/x)² = 2

> (x-1/x) = √2

(x+1/x)² = 6

> (x+1/x) = √6

Now multiplying (x+1/x) with (x-1/x)

> x² - 1/x² = √6 × √2 = √12 = 2√3 .

Answer -

(1) √2

(2) 2√3

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Additional Information -

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

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Answer 2

Step-by-step explanation:

$\leadsto \tt \: {x}^{2} + \frac{1}{ {x}^{2} } = 6$

Subtracting 2 on both sides

$\leadsto \tt \: {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 6 -2 \\ \leadsto \tt(x - \frac{1}{x} ) {}^{2} = 4 \\ \leadsto \tt \: \boxed{ \tt \red{ x - \frac{1}{x} = \sqrt{4} = 2}}$

Adding 2 on both sides

$\leadsto \tt \: {x}^{2} + \frac{1}{ {x}^{2} } = 6$

Adding 2 on both sides

$\tt \leadsto {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 6 + 2 \\ \tt \leadsto(x + \frac{1}{x}) {}^{2} = 8 \\ \tt \leadsto \: x + \frac{1}{x} = \sqrt{8}$

Subtracting 2 on both sides

Multiplying x+1/x and x-1/x, we get

(x+1/x)(x-1/x)=√8 × 2

x²-1/x²=√8×√4

x²-1/x²=√32=4√2