1. Solve the equation $\frac{1}{x} = 2x + 3$. Multiply both sides by $x$ (assuming $x \neq 0$) to get: $$1 = 2x^2 + 3x$$ Rearrange the equation to standard quadratic form: $$2x^2 + 3x - 1 = 0$$ Use the quadratic formula, where $a=2$, $b=3$, and $c=-1$: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$$ So the solutions are: $$x = \frac{-3 + \sqrt{17}}{4}$$ and $$x = \frac{-3 - \sqrt{17}}{4}$$ 2. Consider the function $f(x) = x^2 - 4x + 9$, for all real $x$. Show that $f(x)$ is positive for all $x$. Rewrite $f(x)$ by completing the square: $$f(x) = x^2 - 4x + 9 = (x^2 - 4x + 4) + 5 = (x - 2)^2 + 5$$ Since $(x-2)^2 \geq 0$ for all real $x$, we have $$f(x) \geq 5 > 0$$ Thus, $f(x)$ is positive for all $x \in \mathbb{R}$.