Do perfect square trinomials only have one root? I apologize for the basic question, but I'm just now learning of perfect square trinomials in my math class. Google hasn't provided any relevant answers.
Throughout all of the examples I have been given with perfect square trinomials ($49x^2 - 14x + 1$ and
$9a^2 + 24a + 16$, 
etc.) each has only turned up with one root. Is this true for all perfect square trinomials? Is there an explanation?
 A: It seems that by a trinomial you mean certain polynomials of order two.
I will assume so.
For other kinds of perfect squares (squares of polynomials of order higher than one), see MPW's answer.
A perfect square trinomial is of the form $(ax+b)^2$.
You know that $y^2=0$ is zero if and only if $y=0$, so $(ax+b)^2=0$ if and only if $ax+b=0$.
But this has only one root, right?
In fact, a trinomial is a perfect square if and only if it has only one zero.
A: I don't believe the other answers are correct. There can be multiple roots. There will be a single root if and only if the perfect square trinomial is of degree $2$. For example,
$$x^4 - 2x^3 + x^2 = (x^2 - x)^2$$
is a perfect square trinomial, and it has roots $0$ and $1$ since it can be factored further as $$x^2(x-1)^2$$
A: Yes, they have only one root: Note that being a perfect square means that your trinomial is of the form $\alpha^2 x^2 + 2\alpha\beta x + \beta^2$, and can be written as $(\alpha x +\beta)^2$. Now $0$ is the only real number which has square $0$, that is 
$$ \alpha^2 x^2 + 2\alpha\beta x + \beta^2 = (\alpha x + \beta)^2  = 0 \iff \alpha x + \beta = 0 $$
which is true exactly iff $x = -\frac{\beta}{\alpha}$.
