Quadratic equations greater than zero, then we have at most one real root

Question

I'm watching this video on what's a metric (in topology), specifically I'm at the 5th minute, where the interlocutor says that quadratic equations greater than zero have at most one real root, because it touches zero only once. Specifically, the equation is talking about is $$f(t) = \sum_{i=1}^n (a_i + tb_i)^2 \geq 0$$

I'm aware of when the discriminant of the quadratic formula is negative, $0$ or positive we have respectively no, one and $2$ solutions (because of obvious reasons from arithmetic), but I'm not really getting the statement above.

Maybe it's an easy thing and I'm simply not currently seeing it...

Answer 1

Since we are talking about real (and not complex) quantities, every square is $\ge 0$. Furthermore, if $f(t) = 0$ then all the squares inside $f$ must be $0$, i.e. $(a_i + t b_i)^2 = 0$ for all $i$, meaning that $a_i + t b_i = 0$ for all $i$. This implies $t = - \frac {a_i} {b_i}$ for all $i$. Letting $k$ be the common value of all these fractions, we get $a_i = k b_i$ for all $i$ so $f$ looks like $\sum_i (k b_i + t b_i)^2 = (k+t)^2 \sum_i b_i^2$.

To conclude, either $f$ has no root, or if it has, then $f$ must have the very peculiar form shown above and the root is unique (well, if you really want to be strict, then the root has multiplicity $2$).

Answer 2

$x^2$ for any $x \in \Bbb{R}$ is $0$ if and only if $x=0$. All other times, they are positive. Thus, summing $(a_i+b_i\cdot t)^2$ is summing a bunch of non-negative expressions and that sum can only be $0$ when all of them are zero: That is, they can only be zero when all of the $a_i+b_i\cdot t=0$, which can happen at most for one value of $t$.

For example, the following has one solution: $$(t+2)^2 = 0$$ The above is only true when $t=-2$ since that's when $t+2=0$. All other times, the left-hand side is positive. Here, the discriminant is $0$.

Also, the following has no solution: $$(t+2)^2+(3t+1)^2=0$$ This has no solutions because for the first term to be $0$, we need $t=-2$, but for the second term to be $0$, we need $t=-\frac{1}{3}$, meaning that both can not be $0$ at the same time, so the left-hand side is always positive. Here, the discriminant is $-100$.

Finally, the following has one solution: $$(t+2)^2+(2t+4)^2=0$$ The first term is $0$ when $t+2=0 \implies t=-2$ and the second term is also $0$ when $2t+4=0 \implies t=-2$, so we get that there is one solution of $t=-2$ and for all other $t$, the left-hand side is positive. Here, the discriminant is $0$.