Question

If $x^4+\frac{1}{x^4}=623$, then $x+\frac{1}{x}=$

• A. $3\sqrt{3}$
• B. $25$
• C. $27$
• D. $-3\sqrt{3}$

Answer 1

Answer: (A), (D)

$-3\sqrt{3}$

Given: $x^4+\frac{1}{x^4}=623$
$\therefore x^4+\frac{1}{x^4}+2=623+2$
${(x^2+\frac{1}{x^2})}^2=625$
$(x^2+\frac{1}{x^2})=25$
$[\because \text{Sum of squares of real number is always positive}]$
$\therefore (x^2+\frac{1}{x^2}+2)=25+2$
${(x+\frac{1}{x})}^2=27$
$x+\frac{1}{x}=\pm \sqrt{3\times 9}$
$x+\frac{1}{x}=\pm 3\sqrt{3}$

Hence, $\left(c\right)$ and $\left(d\right)$ are the correct options.