How to quickly tell that a quadratic only has a single root?

Question

Hello good math wizards, I'm trying to figure out why the following equation has at most one root:

$$f (t) = \textbf{x} \cdot \textbf{x} + \textbf{x} \cdot t\textbf{y} + t\textbf{y} \cdot \textbf{x} + t\textbf{y} \cdot t\textbf{y} = \left \| \textbf{x} \right \|^2 + 2(\textbf{x} \cdot \textbf{y})t + \left \| \textbf{y} \right \|^2t^2$$

where $\textit{t}$ is a real number, $\textbf{x}$ and $\textbf{y}$ are fixed vectors in $\mathbb{R}^n$, and $\textbf{y}\neq0$

The background of this question is that I am going through the text here, and in the section about the Cauchy-Schwarz Inequality the authors state "Hence f is a quadratic polynomial with at most one root." However, I don't understand why this quadratic has at most one root. Can someone help clarify why it should be immediately apparent that his equation has only one root?

Just FYI this isn't homework or anything; I'm just going through the text. Thanks in advance!

Answer 1

Notice that $$f(t) = ||\mathbf{x}+t\mathbf{y}||^2$$ Thus $f$ can obviously only have one root, which would be $\mathbf{x}+t\mathbf{y} = 0$.

Answer 2

The "hence" is referring back to the fact that $f(t) \geq 0$, from the previous sentence. A quadratic polynomial in one variable has a unique extremum, so if it's never negative it can have at most one root.