If $a$ is a constant such that $4x^2 + 14x + a$ is the square of a binomial, then what is $a$?
If $4x^2 + 14x + a$  is the square of a binomial, then the binomial has the form $2x +b$ for some number $b$, because $(2x)^2 = 4x^2$.  So, we compare $(2x+b)^2$ to $4x^2 + 14x + a$. Expanding $(2x+b)^2$ gives \[(2x+b)^2 = (2x)^2 + 2(2x)(b) + b^2 = 4x^2 + 4bx + b^2.\] Equating the linear term of this to the linear term of $4x^2+14x+a$, we have $4bx=14x$, so $b=\frac{14}{4}=\frac{7}{2}$. Squaring the binomial gives $\left(2x+\frac{7}{2}\right)^2=4x^2+14x+\frac{49}{4}$. Therefore, $a=\boxed{\frac{49}{4}}$.