If \(\frac{8x}{2x^2 + 7x - 2}\) = 1, where x > 0, then find x2 + \(\frac{1}{x^2}\).
Answer & explanation
Correct answer: option 4
8x = 2x2 + 7x - 2
2x2 - x - 2 = 0
2x (x - \(\frac{1}{2}\) - \(\frac{1}{x}\)) = 0
x - \(\frac{1}{x}\) = \(\frac{1}{2}\)
x2 + \(\frac{1}{x^2}\) = (\(\frac{1}{2}\))2 + 2 = \(\frac{9}{4}\)