Discussion

If \(x^{2} + x - 1 = 0\) what is the value of \(x^{4} + \frac{1}{x^{4}}\)?

Answer: C

The value is 7.

Divide \(x^{2} + x - 1 = 0\) by \(x\) (\(x \ne 0\)) to get \(x + 1 - \frac{1}{x} = 0\), so \(x - \frac{1}{x} = -1\). Square twice:

\[ \begin{aligned} x^{2} + \frac{1}{x^{2}} &= \left(x - \frac{1}{x}\right)^{2} + 2 = 3 \\ x^{4} + \frac{1}{x^{4}} &= 3^{2} - 2 = 7 \end{aligned} \]

9 forgets to subtract 2 in the last step, and 1 and 5 do not follow from the given equation.

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