Question

Pls armies help me with this​

Answer 1

Answer :

The value of $\sf{x-\frac{1}{x} \text { is } 5}$

To find:

$\sf \:x-\frac{1}{x}$

Detailed Solution:

Given:

$\sf\:x^{2}+\frac{1}{x^{2}}=27$

We know that $\sf\:(a-b)^{2}=a^{2}+b^{2}-2 a b$

Substitute a = x, b = $\sf\frac{1}{x}$

$\sf\implies\:(x-\frac{1}{x})^{2}$

= $x^{2}+\frac{1}{x^{2}}-2 \times x \times \frac{1}{x}$

= $x^{2}+\frac{1}{x^{2}}-2$

=27-2 ( It is given that $\sf\:x^{2}+\frac{1}{x^{2}}=27$)

Thus ,

• $\sf(x-\frac{1}{x})^{2}$

$\sf(x-\frac{1}{x})^{2}=(5)^{2}$

Take the square root of LHS and RHS ,

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