If x8 - 1442x4 + 1 = 0, then a possible value of $$x - \frac{1}{x}$$ is:
A. 5
B. 4
C. 6
D. 8
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {x^8} - 1442{x^4} + 1 = 0 \cr & \frac{{{x^8}}}{{{x^4}}} - \frac{{1442{x^4}}}{{{x^4}}} + \frac{1}{{{x^4}}} = 0 \cr & {x^4} - 1442 + \frac{1}{{{x^4}}} = 0 \cr & {x^4} + \frac{1}{{{x^4}}} = 1442 \cr & {x^4} + \frac{1}{{{x^4}}} + 2 = 1444 \cr & \left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 38 \cr & {x^2} + \frac{1}{{{x^2}}} - 2 = 36 \cr & {\left( {x - \frac{1}{x}} \right)^2} = 36 \cr & x - \frac{1}{x} = 6 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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