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

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 ${(\frac{x^2-c^2}{\sqrt{x^2-c^2}})}^2={39}^2$, or $\frac{(x^2-c^2)(x^2-c^2)}{x^2-c^2}={39}^2$. 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 $\frac{(x^2-c^2)(x^2-c^2)}{x^2-c^2}={39}^2$ can be rewritten as $(\frac{x^2-c^2}{x^2-c^2})(x^2-c^2)={39}^2$, or $(1)(x^2-c^2)={39}^2$, 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.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.