I know that the square of a binomial has the form $(x+a)^2=x^2+2ax+a^2$ for some constant $a$. So I want to find a value of $c$ that makes $x^2+cx+9c$ match this form. That means I need to have $c=2a$ and $9c=a^2$. I can substitute $c=2a$ into the second equation and get $18a=a^2$. I can solve this for $a$ by subtracting $18a$ from both sides and factoring: $a^2-18a=0 \implies a(a-18)=0$. So $a$ is either $0$ or $18$. But if $a=0$, then $c=0$, which contradicts the assumption that $c$ is nonzero. So $a$ must be $18$. That means $c=2a=2(18)=36$.


If $c$ is a nonzero constant such that $x^2+cx+9c$ is equal to the square of a binomial, then what is $c$?