Question

If x 2 +1/x 2 =27, find x-1/x.

Answer 1

[tex] x^{2} + \frac{1}{ x^{2} } = 27 x^{2} + \frac{1}{ x^{2} } + 2 (x) ( \frac{1}{x}) - 2 (x)( \frac{1}{x} ) = 27 x^{2} + ( \frac{1}{ x^{2} } + 2 ( x - \frac{1}{x} = 27 (x - 1/ x )^{2} = 27 -2 ( x - 1/x)^{2} + \sqrt{25} x - 1/x = + or - 5 [/tex]

Answer 2

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

To find:

Find $x-\frac{1}{x}$

Solution:

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

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

Putting $a=x, b=\frac{1}{x}$

$\left(x-\frac{1}{x}\right)^{2}$

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

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

$=27-2\ (Given\ that\ x^{2}+\frac{1}{x^{2}}=27)$

$=25$  

Hence, $\left(x-\frac{1}{x}\right)^{2}=25$

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

Taking square root of both sides, we get

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