Question

if $x^{2}+\frac{1}{x^{2}}=27$ find the value of $x-\frac{1}{x}$

Answer 1

$\huge\underline\red{Answer:-}$

$x-\frac{1}{x} = ±5$

$\huge\underline\green{Given:-}$

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

$\huge\underline\pink{To \ find:-}$

$The \ value \ of \ x-\frac{1}{x}$

$\huge\underline\orange{Step \ by \ step \ explanation:-}$

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

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

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

$→(x-\frac{1}{x})^{2}=27-2$  $[x^{2}+\frac{1}{x^{2}}=27(given)]$

$→(x-\frac{1}{x})^{2}=25$

$→(x-\frac{1}{x})^{2}=(±5)^{2}$

$\boxed{x-\frac{1}{x}=±5}$

Answer 2

Step-by-step explanation:

x-x/1 =5 is your answer Please follow me permanently and take 10 thanks in return