Question:
If
\[ x^4+\frac{1}{x^4}=623, \] then \[ x+\frac{1}{x}= \]
(a) 27
(b) 25
(c) \[ 3\sqrt{3} \]
(d) \[ -3\sqrt{3} \]
Solution:
Using identity:
\[ x^4+\frac{1}{x^4} = \left(x^2+\frac{1}{x^2}\right)^2-2 \]
Substituting the given value:
\[ 623 = \left(x^2+\frac{1}{x^2}\right)^2-2 \]
\[ \left(x^2+\frac{1}{x^2}\right)^2 = 625 \]
\[ x^2+\frac{1}{x^2} = 25 \]
Now using:
\[ x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2-2 \]
\[ 25 = \left(x+\frac{1}{x}\right)^2-2 \]
\[ \left(x+\frac{1}{x}\right)^2 = 27 \]
\[ x+\frac{1}{x} = \sqrt{27} \]
\[ =3\sqrt{3} \]
Hence, the correct answer is:
\[ \boxed{3\sqrt{3}} \]