SAT Question 7355b9d9: Equivalent Expressions answer and explanation with Desmos solution

Choice D is correct. If $k-x$ is a factor of the expression $-x^2+\left(\frac{1}{29}\right)n{k}^2$, then the expression can be written as $\left(k-x\right)\left(ax+b\right)$, where $a$ and $b$ are constants. This expression can be rewritten as $akx+bk-a{x}^2-bx$, or $-a{x}^2+\left(ak-b\right)x+bk$. Since this expression is equivalent to $-x^2+\left(\frac{1}{29}\right)n{k}^2$, it follows that $-a=-1$, $ak-b=0$, and $bk=\left(\frac{1}{29}\right)n{k}^2$. Dividing each side of the equation $-a=-1$ by $-1$ yields $a=1$. Substituting $1$ for $a$ in the equation $ak-b=0$ yields $k-b=0$. Adding $b$ to each side of this equation yields $k=b$. Substituting $k$ for $b$ in the equation $bk=\left(\frac{1}{29}\right)n{k}^2$ yields $k^2=\left(\frac{1}{29}\right)n{k}^2$. Since $k$ is positive, dividing each side of this equation by $k^2$ yields $1=\left(\frac{1}{29}\right)n$. Multiplying each side of this equation by $29$ yields $29=n$.

Alternate approach: The expression $x^2-y^2$ can be written as $\left(x-y\right)\left(x+y\right)$, which is a difference of two squares. It follows that $\left(\frac{1}{29}\right)n{k}^2-x^2$ is equivalent to $\left(\left(\sqrt{\frac{1}{29}n}\right)k-x\right)\left(\left(\sqrt{\frac{1}{29}n}\right)k+x\right)$. It's given that $k-x$ is a factor of $-x^2+\left(\frac{1}{29}\right)n{k}^2$, so the factor $\left(\sqrt{\frac{1}{29}n}\right)k-x$ is equal to $k-x$. Adding $x$ to both sides of the equation $\left(\sqrt{\frac{1}{29}n}\right)k-x=k-x$ yields $\left(\sqrt{\frac{1}{29}n}\right)k=k$. Since $k$ is positive, dividing both sides of this equation by $k$ yields $\sqrt{\frac{1}{29}n}=1$. Squaring both sides of this equation yields $\frac{1}{29}n=1$. Multiplying both sides of this equation by $29$ yields $n=29$.