Question 117·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

No Desmos needed. The product of a quadratic’s solutions comes from its coefficients. The leading coefficient is the number multiplying $x^2$. Only the $x$ terms on the left multiply to make an $x^2$ term:

The right side has no $x^2$, so the leading coefficient is $ab$ when the equation is set equal to $0$.

The left side has constant term $b\cdot a=ab$. The right side, $57ab(x+1)$, has constant term $57ab$. Keep both constants: subtract the one on the right to set the equation equal to $0$:

So its constant term is $-56ab$, not $-57ab$.

For $Ax^2+Bx+C=0$, the product of the solutions is $C/A$. Put in the constant term and leading coefficient:

Since $a$ and $b$ are positive, $ab$ isn’t $0$. Cancel it:

The product is $-56$ for every allowed $a$ and $b$. Choice D.

The problem says $k$ does not depend on $a$ or $b$. Since $1$ is positive, you can set $a=b=1$ and get the same product as in any allowed case. Substitute those values into the equation:

Type both sides as graphs: $y=(x+1)^2$ and $y=57(x+1)$. Click their intersections. Desmos shows $(-1,0)$ and $(56,3249)$, so the solutions are the $x$-coordinates $-1$ and $56$. Don’t divide by $x+1$: it equals $0$ at $x=-1$, and division would lose that solution.

Multiply both solutions, not just the positive one:

Because the product does not depend on $a$ or $b$, it is $-56$ in the original equation too. Choice D.