SAT Question 2c05d312: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice A is correct. The left-hand side of the given equation is the expression $57x^2+\left(57b+a\right)x+ab$. Applying the distributive property to this expression yields $57x^2+57bx+ax+ab$. Since the first two terms of this expression have a common factor of $57x$ and the last two terms of this expression have a common factor of $a$, this expression can be rewritten as $57x\left(x+b\right)+a\left(x+b\right)$. Since the two terms of this expression have a common factor of $\left(x+b\right)$, it can be rewritten as $\left(x+b\right)\left(57x+a\right)$. Therefore, the given equation can be rewritten as $\left(x+b\right)\left(57x+a\right)=0$. By the zero product property, it follows that $x+b=0$ or $57x+a=0$. Subtracting $b$ from both sides of the equation $x+b=0$ yields $x=-b$. Subtracting $a$ from both sides of the equation $57x+a=0$ yields $57x=-a$. Dividing both sides of this equation by $57$ yields $x=\frac{-a}{57}$. Therefore, the solutions to the given equation are $-b$ and $\frac{-a}{57}$. It follows that the product of the solutions of the given equation is $\left(-b\right)\left(\frac{-a}{57}\right)$, or $\frac{ab}{57}$. It's given that the product of the solutions of the given equation is $kab$. It follows that $\frac{ab}{57}=kab$, which can also be written as $ab\left(\frac{1}{57}\right)=ab\left(k\right)$. It's given that $a$ and $b$ are positive constants. Therefore, dividing both sides of the equation $ab\left(\frac{1}{57}\right)=ab\left(k\right)$ by $ab$ yields $\frac{1}{57}=k$. Thus, the value of $k$ is $\frac{1}{57}$.