Question

if x^2+ 1/x^2=66 find the value of x-1/x

Answer 1

Hey here is your answer
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$\\ {x}^{2} + \frac{1}{ {x}^{2} } = 66 \\ \\ now \: . \\ {(x - \frac{1}{x}) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \times x \times \frac{1}{x} \\ \\ {(x - \frac{1}{x}) }^{2} =66 - 2 \\ \\ {(x - \frac{1}{x}) }^{2} =64 \\ \\ {(x - \frac{1}{x}) }^{2} = {8}^{2} \\ \\ x - \frac{1}{x} = 8$
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Answer 2

The value of  $x-\dfrac{1}{x}=8$.

Step-by-step explanation:

We have,

$x^{2} +\dfrac{1}{x^{2}} =66$

To find, the value of $x-\dfrac{1}{x} =?$

We know that,

$(x-\dfrac{1}{x})^{2}=x^{2} +\dfrac{1}{x^{2}} -2.x.\dfrac{1}{x}$

⇒ $(x-\dfrac{1}{x})^{2}=x^{2} +\dfrac{1}{x^{2}} -2$

⇒ $(x-\dfrac{1}{x})^{2}=66 -2=64$

⇒ $(x-\dfrac{1}{x})^{2}=8^{2}$

⇒ $x-\dfrac{1}{x}=8$

Hence, the value of  $x-\dfrac{1}{x}=8$.