If (x² + \(\frac{1}{x^2}\) - a)2 + (x + \(\frac{1}{x}\) - b)2 = 0, where a and b are real numbers and x ≠ 0, then find the value of a in terms of b.
Answer & explanation
Correct answer: option 4
(x² + \(\frac{1}{x^2}\) - a)2 + (x + \(\frac{1}{x}\) - b)2 = 0
⇒ (x² + \(\frac{1}{x^2}\) - a)2 = 0
⇒ x² + \(\frac{1}{x^2}\) = a .... (i)
and
⇒ (x + \(\frac{1}{x}\) - b)2 = 0
⇒ x + \(\frac{1}{x}\) = b
squaring both sides →
x² + \(\frac{1}{x^2}\) + 2 = b2
a + 2 = b² (using ... (i))
a = b² - 2