Question

Choose the correct answer from the alternative given.
If $x^4+\frac{1}{x^4}=14$ then find the value of $x-\frac{1}{x}$

• A. $\pm \sqrt{2}$
• B. $\sqrt{12}$
• C. $\pm \sqrt{6}$
• D. None of these

Answer 1

Answer: (A)

$\pm \sqrt{2}$

$x^4+\frac{1}{x^4}$ $=14$
Adding 2 on both sides,$x^4+\frac{1}{x^4}$ $+2=14+2$
${(x^2+\frac{1}{x^2})}^2=16\Rightarrow x^2+\frac{1}{x^2}=\pm 4$
Neglecting the negative sign, as sum of squares of two real numbers cannot be negative.
$x^2+\frac{1}{x^2}$ $=4$
Substracting 2 from both the sides, $x^2+\frac{1}{x^2}-2=4-2\Rightarrow {(x-\frac{1}{x})}^2=2\Rightarrow x-\frac{1}{x}=\pm \sqrt{2}$