Nonlinear equations in one variable and systems of equations in two variables SAT question

The correct answer is $\frac{1}{16}$. Let $p$ and $q$ represent the solutions to the given equation. Then, the given equation can be rewritten as $64(x-p)(x-q)=0$, or $64x^2-64(p+q)+pq=0$. Since this equation is equivalent to the given equation, it follows that $-(16a+4b)=-64(p+q)$. Dividing both sides of this equation by $-64$ yields $\frac{16a+4b}{64}=p+q$, or $\frac{1}{16}(4a+b)=p+q$. Therefore, the sum of the solutions to the given equation, $p+q$, is equal to $\frac{1}{16}(4a+b)$. Since it's given that the sum of the solutions to the given equation is $k(4a+b)$, where $k$ is a constant, it follows that $k=\frac{1}{16}$. Note that 1/16, .0625, 0.062, and 0.063 are examples of ways to enter a correct answer.

Alternate approach: The given equation can be rewritten as $64x^2-4(4a+b)x+ab=0$, where $a$ and $b$ are positive constants. Dividing both sides of this equation by $4$ yields $16x^2-(4a+b)x+\frac{ab}{4}=0$. The solutions for a quadratic equation in the form $A{x}^2+Bx+C=0$, where $A$, $B$, and $C$ are constants, can be calculated using the quadratic formula, $x=\frac{-B+\sqrt{B^2-4AC}}{2A}$ and $x=\frac{-B-\sqrt{B^2-4AC}}{2A}$. It follows that the sum of the solutions to a quadratic equation in the form $A{x}^2+Bx+C=0$ is $\frac{-B+\sqrt{B^2-4AC}}{2A}+\frac{-B-\sqrt{B^2-4AC}}{2A}$, which can be rewritten as $\frac{-B+-B+\sqrt{B^2-4AC}-\sqrt{B^2-4AC}}{2A}$, which is equivalent to $\frac{-2B}{2A}$, or $-\frac{B}{A}$. In the equation $16x^2-(4a+b)x+\frac{ab}{4}=0$, $A=16$, $B=-(4a+b)$, and $C=\frac{ab}{4}$. Substituting $16$ for $A$ and $-(4a+b)$ for $B$ in $-\frac{B}{A}$ yields $-\frac{-(4a+b)}{16}$, which can be rewritten as $\frac{1}{16}(4a+b)$. Thus, the sum of the solutions to the given equation is $\frac{1}{16}(4a+b)$. Since it's given that the sum of the solutions to the given equation is $k(4a+b)$, where $k$ is a constant, it follows that $k=\frac{1}{16}$.