Discussion
If \(x^{4}+\frac{1}{x^{4}}=14\), then the value of \(x-\frac{1}{x}\) is :
Answer: A
The value is \(\sqrt{2}\).
Work down from the fourth powers using the identity \((x \pm \tfrac{1}{x})^{2} = x^{2} + \tfrac{1}{x^{2}} \pm 2\):
\[ \begin{aligned} \left(x^{2} + \tfrac{1}{x^{2}}\right)^{2} &= 14 + 2 = 16 \\ x^{2} + \tfrac{1}{x^{2}} &= 4 \\ \left(x - \tfrac{1}{x}\right)^{2} &= 4 - 2 = 2 \end{aligned} \]
So \(x - \tfrac{1}{x} = \sqrt{2}\), taking the positive root as the options do.
\(\sqrt{3}\), \(\sqrt{5}\) and \(\sqrt{7}\) do not square to 2. Note that \(x + \tfrac{1}{x}\) would be \(\sqrt{6}\), since its square is 4 + 2; the question asks for the difference, not the sum.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development