Question

$\huge \: if \: {x}^{2} + \frac{1}{ {x}^{2} } = 102$
$find \: \: \: the \: \: \: value \: \: \: of \: - -$
$\huge \: x - \frac{1}{x}$
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Answer 1

Step-by-step explanation:

mention value of x and y as mentioned

if so, put the formula

let x be X and 1/x be y

x²+1/x²=(x-y)²+2xy

=102²+2× value of x × y

thank you hope it helps you

Answer 2

Answer:

The value of x - 1/x is ±10

Step-by-step explanation:

Given :

$\sf x^2 + \dfrac{1}{x^2}=102$

To find :

the value of x - 1/x

Solution :

We know that,

(a - b)² = a² + b² - 2ab

Put a = x and b = 1/x,

$\sf \bigg(x-\dfrac{1}{x} \bigg)^2 = x^2 + \dfrac{1}{x^2} - 2(x) \bigg(\dfrac{1}{x}\bigg) \\\\ \sf \bigg(x-\dfrac{1}{x} \bigg)^2= 102 - 2 \\\\ \sf \bigg(x-\dfrac{1}{x} \bigg)^2 = 100 \\\\ \sf \bigg(x-\dfrac{1}{x} \bigg) = \sqrt{100} \\\\ \sf x-\dfrac{1}{x} = \pm 10$

Therefore, the value of x - 1/x is ±10