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If \(x^8 + \frac{1}{x^8} = 2207\), what could be the value of \(x + \frac{1}{x}\) ?
I. -3
II. 1
III. 3
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
Are You Up For the Challenge: 700 Level Questions
I. -3
II. 1
III. 3
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
Are You Up For the Challenge: 700 Level Questions
29 Nov 2020, 23:54
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CEdward
Hello CEdward. Here we should know the expansion of \((a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2}+ 2\), which is derived from the expansion of \((a + b)^2 = a^2 + b^2 + 2ab\).
Therefore \(a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2\)
Now \(x^8 + \frac{1}{x^8} = (x^4 + \frac{1}{x^4})^2 - 2 = 2207\)
\((x^4 + \frac{1}{x^4})^2 = 2209\)
Now to find the square root of 2209, we know that \(50^2\) is 2500 and \(45^2\) (there is a short cut to find squares of numbers ending in 5) is 2025..So the square root should lie between between 45 and 50 and the square should end with 9. S0 \(47^2\) fits that.
\(x^4 + \frac{1}{x^4} = 47\)
\((x^2 + \frac{1}{x^2})^2 - 2 = 47\)
\((x^2 + \frac{1}{x^2})^2 = 49\)
\(x^2 + \frac{1}{x^2} = 7\)
Now, \((x + \frac{1}{x})^2 - 2 = 7\)
\((x + \frac{1}{x})^2 = 9\)
\(x + \frac{1}{x} = \pm 3\)
Hope this helps
Arun Kumar
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