Find $a$ such that $ax^2+15x+4$ is the square of a binomial.
I want to find a value of $a$ such that $ax^2+15x+4$ is the square of a binomial of the form $bx+c$. I can expand the left-hand side of this equation and compare the coefficients of $x^2$, $x$, and the constant terms. Expanding the left-hand side, I get $(bx+c)^2 = b^2x^2 + 2bcx + c^2$. Comparing the coefficients of $x^2$, I see that $a = b^2$. Comparing the coefficients of $x$, I see that $15 = 2bc$. Comparing the constant terms, I see that $4 = c^2$. Since $c^2 = 4$, I can conclude that $c = \pm 2$. If $c = 2$, then $15 = 2b \cdot 2$, so $b = \frac{15}{4}$. If $c = -2$, then $15 = 2b \cdot (-2)$, so $b = -\frac{15}{4}$. In either case, $b^2 = \frac{225}{16}$. Therefore, $a = b^2 = \frac{225}{16}$. 