Question

if X + 1 by x is equal to 6 find x square minus one by x square​

Answer 1

Given :-

• (x + 1/x) = 6

To Find :-

• (x² - 1/x)² = ?

Solution :-

→ (x + 1/x) = 6

Squaring Both Sides we get,

→ (x + 1/x)² = 6²

→ x² + 1/x² + 2 * x * 1/x = 36

→ x ² + 1/x² = 36 - 2

→ x² + 1/x² = 34

Subtracting 2 both sides now,

→ x² + 1/x² - 2 = 34 - 2

→ x² + 1/x² - 2 * x * 1/x = 32

→ (x - 1/x)² = 32

→ (x - 1/x) = √32 = ±4√2.

_______________________

So,

→ (x² - 1/x²)

→ (x + 1/x)(x - 1/x)

→ 6 * (±4√2)

→ ±24√2 (Ans).

Answer 2

Question:

$x + \frac{1}{x} = 6 \: then\: find \: {x}^{2} - \frac{1}{x}$

Answer:

Given

$x + \frac{1}{x} = 6 \\ \\ squaring \: both \: side \\ \\ {(x + \frac{1}{x} )}^{2} = ( {6})^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2x \frac{1}{x} = 36 \\ \\ {x}^{2} + \frac{1}{x} + 2 = 36 \\ \\ {x}^{2} + \frac{1}{x} = 36 - 2 \\ \\ {x}^{2} + \frac{1}{x} = 34$

Now ,sub 2 both side

${x}^{2} + \frac{1}{ {x}^{2} } - 2 = 34 - 2 \\ \\ \because \: ( {x} - \frac{1}{ {x}} )^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ ({x} -\frac{1}{ {x}} )^{2} = 32 \\ \\ {x} - \frac{1}{ {x}} = \sqrt{32}$

Then,

$(x + \frac{1}{x} )(x - \frac{1}{x}) = {x}^{2} - \frac{1}{ {x}^{2} } \\ \\ (6) \times \sqrt{32} = {x}^{2} - \frac{1}{ {x}^{2} } \\ \\ 6 \sqrt{32} = {x}^{2} - \frac{1}{ {x}^{2} }$