SAT Question 104bff62: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice D is correct. If $x^2-c^2\le 0$, then neither side of the given equation is defined and there can be no solution. Therefore, $x^2-c^2>0$. Subtracting $\frac{c^2}{\sqrt{x^2-c^2}}$ from both sides of the given equation yields $\frac{x^2}{\sqrt{x^2-c^2}}-\frac{c^2}{\sqrt{x^2-c^2}}=39$, or $\frac{x^2-c^2}{\sqrt{x^2-c^2}}=39$. Squaring both sides of this equation yields ${\left(\frac{x^2-c^2}{\sqrt{x^2-c^2}}\right)}^2=39^2$, or $\frac{\left(x^2-c^2\right)\left(x^2-c^2\right)}{x^2-c^2}=39^2$. Since $x^2-c^2$ is positive and, therefore, nonzero, the expression $\frac{x^2-c^2}{x^2-c^2}$ is defined and equivalent to $1$. It follows that the equation $\frac{\left(x^2-c^2\right)\left(x^2-c^2\right)}{x^2-c^2}=39^2$ can be rewritten as $\left(\frac{x^2-c^2}{x^2-c^2}\right)\left(x^2-c^2\right)=39^2$, or $\left(1\right)\left(x^2-c^2\right)=39^2$, which is equivalent to $x^2-c^2=39^2$. Adding $c^2$ to both sides of this equation yields $x^2=c^2+39^2$. Taking the square root of both sides of this equation yields two solutions: $x=\sqrt{c^2+39^2}$ and $x=-\sqrt{c^2+39^2}$. Therefore, of the given choices, $-\sqrt{c^2+39^2}$ is one of the solutions to the given equation.