# 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?

• if you put a dollar sign at the front and end of each equation it will make them pretty and easier to follow – scrappedcola Jul 7 '15 at 13:40
• Your polynomial is of the form $(a x+b)^2$? How many roots does this have? – David Mitra Jul 7 '15 at 13:40
• Thanks everyone for the help! – Connor Smith Jul 7 '15 at 14:58

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.

• I don't think this is correct. See my answer. – MPW Jul 7 '15 at 13:55
• @MPW, thanks. I added a note. I made an implicit assumption explicit now. – Joonas Ilmavirta Jul 7 '15 at 13:58
• Okay, but "trinomial" means "polynomial with three nonzero terms" and has no connection with the degree of the polynomial – MPW Jul 7 '15 at 13:59
• @MPW, true. But it seemed to me that the OP was only interested in second order polynomials that happen to be perfect squares, so I confined myself to those. I should have made the remark in the first version, though. – Joonas Ilmavirta Jul 7 '15 at 14:00

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 $$x^2(x-1)^2$$

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 $x = -\frac{\beta}{\alpha}$.

• Is $x^4 - 2x^2 + 1$ not a "perfect square trinomial"? – John Brevik Jul 7 '15 at 13:50