Find $a$ such that $ax^2+15x+4$ is the square of a binomial.
The square of the binomial $rx+s$ is  \[(rx+s)^2=r^2x^2+2rsx+s^2.\]If this is equal to $ax^2+15x+4$, then $s$ must be either 2 or -2.  Since $(rx+s)^2=(-rx-s)^2$, we may choose either $s=2$ or $s=-2$, and the solution will be the same.  We choose $s=2$.

The square of $rx+2$ is  \[(rx+2)^2=r^2x^2+4rx+4.\]If this is equal to $ax^2+15x+4$ then we must have $15=4r$ or $r=\frac{15}4$.  This gives our square: \[\left(\frac{15}4x+2\right)^2=\frac{225}{16}x^2+15x+4.\]Therefore $a=\boxed{\frac{225}{16}}$.
\frac{225}{16}
Level 5
Algebra
Problem: Find $a$ such that $ax^2+15x+4$ is the square of a binomial.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the value of $a$, we need to express the quadratic $ax^2 + 15x + 4$ as the square of a binomial, which takes the form $(bx + c)^2$. Expanding this binomial gives us $b^2x^2 + 2bcx + c^2$. 

By comparing coefficients, we can set up the following equations:
1. From the coefficient of $x^2$: $b^2 = a$.
2. From the coefficient of $x$: $2bc = 15$.
3. From the constant term: $c^2 = 4$.

From the third equation, we find that $c$ can be either $2$ or $-2$. 

First, consider $c = 2$:
Substituting $c = 2$ into the second equation gives $2b(2) = 15$, leading to $4b = 15$, thus $b = \frac{15}{4}$. Now substituting $b$ back into the first equation gives:
$$ a = b^2 = \left(\frac{15}{4}\right)^2 = \frac{225}{16}. $$

Next, consider $c = -2$:
Substituting $c = -2$ into the second equation gives $2b(-2) = 15$, leading to $-4b = 15$, thus $b = -\frac{15}{4}$. Substituting this back into the first equation gives:
$$ a = b^2 = \left(-\frac{15}{4}\right)^2 = \frac{225}{16}. $$

In both cases, we arrive at the same value for $a$. Therefore, the final answer is $\frac{225}{16}$.<|END_OF_LATENT|> The final answer is $\frac{225}{16}$.