jmc

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, $a=\boxed{\frac{49}{4}}$.