Question: Solve for $x$.
$1 = x - \frac{20}{x}$
1. **State the problem:** Solve the equation $$1 = x - \frac{20}{x}$$ for $x$.
2. **Rewrite the equation:** Multiply both sides by $x$ to eliminate the fraction:
$$x \cdot 1 = x \cdot \left(x - \frac{20}{x}\right)$$
which simplifies to
$$x = x^2 - 20$$
3. **Rearrange the equation:** Move all terms to one side:
$$0 = x^2 - x - 20$$
4. **Solve the quadratic equation:** Use the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a=1$, $b=-1$, and $c=-20$.
Calculate the discriminant:
$$\Delta = (-1)^2 - 4 \cdot 1 \cdot (-20) = 1 + 80 = 81$$
5. **Find the roots:**
$$x = \frac{-(-1) \pm \sqrt{81}}{2 \cdot 1} = \frac{1 \pm 9}{2}$$
6. **Calculate each solution:**
- $$x = \frac{1 + 9}{2} = \frac{10}{2} = 5$$
- $$x = \frac{1 - 9}{2} = \frac{-8}{2} = -4$$
7. **Check for extraneous solutions:** Substitute back into the original equation:
- For $x=5$:
$$1 \stackrel{?}{=} 5 - \frac{20}{5} = 5 - 4 = 1$$ (True)
- For $x=-4$:
$$1 \stackrel{?}{=} -4 - \frac{20}{-4} = -4 + 5 = 1$$ (True)
**Both solutions are valid.**
**Final answer:** $$x = 5, -4$$