Solving for $x^4 + \frac{1}{x^4}$

We are given the equation $x^2 + x - 1 = 0$. Our goal is to find the value of $x^4 + \frac{1}{x^4}$.

Step-by-Step Solution

• Consider the equation $x^2 + x - 1 = 0$. Notice that $x \ne 0$ because substituting $x=0$ gives $-1 = 0$, which is false. We can divide the equation by $x$:

  $ \frac{x^2}{x} + \frac{x}{x} - \frac{1}{x} = \frac{0}{x} $

  This simplifies to:

  $ x + 1 - \frac{1}{x} = 0 $

• Rearrange the terms to isolate $x - \frac{1}{x}$:

  $ x - \frac{1}{x} = -1 $

• Square both sides of the equation $x - \frac{1}{x} = -1$ to find $x^2 + \frac{1}{x^2}$:

  $ \left(x - \frac{1}{x}\right)^2 = (-1)^2 $

  Using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$:

  $ x^2 - 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 1 $

  $ x^2 - 2 + \frac{1}{x^2} = 1 $

  Now, solve for $x^2 + \frac{1}{x^2}$:

  $ x^2 + \frac{1}{x^2} = 1 + 2 $

  $ x^2 + \frac{1}{x^2} = 3 $

• Square both sides of the equation $x^2 + \frac{1}{x^2} = 3$ to find $x^4 + \frac{1}{x^4}$:

  $ \left(x^2 + \frac{1}{x^2}\right)^2 = 3^2 $

  Using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$:

  $ (x^2)^2 + 2(x^2)\left(\frac{1}{x^2}\right) + \left(\frac{1}{x^2}\right)^2 = 9 $

  $ x^4 + 2 + \frac{1}{x^4} = 9 $

  Finally, solve for $x^4 + \frac{1}{x^4}$:

  $ x^4 + \frac{1}{x^4} = 9 - 2 $

  $ x^4 + \frac{1}{x^4} = 7 $

Final Answer Verification

The calculated value of $x^4 + \frac{1}{x^4}$ is 7, which corresponds to Option C.