\(\frac { x^ 2} { \sqrt { x^ 2- c^ 2} } = \frac { c^ 2} { \sqrt { x^ 2- c^ 2} } + 39\) In the given equation, c is a positive constant. Which of the following is one of the solutions to the given equation?A. \(- c\) B. \(- c^ 2- 39^ 2\) C. \(- \sqrt { 39^ 2- c^ 2} \) D. \(- \sqrt { c^ 2+ 39^ 2} \) 

To solve the given equation:
\[\frac { x^ 2} { \sqrt { x^ 2 - c^ 2} } = \frac { c^ 2} { \sqrt { x^ 2 - c^ 2} } + 39\]
First, we can simplify the equation by multiplying both sides by \(\sqrt { x^ 2 - c^ 2} \):
\[x^ 2 = c^ 2 + 39\sqrt { x^ 2 - c^ 2} \]
Next, isolate the square root term:
\[x^ 2 - c^ 2 = 39\sqrt { x^ 2 - c^ 2} \]
Let \(y = \sqrt { x^ 2 - c^ 2} \). Then, \(y^ 2 = x^ 2 - c^ 2\), so the equation becomes:
\[y^ 2 = 39y\]
This simplifies to:
\[y^ 2 - 39y = 0\]
Factor out \(y\):
\[y( y - 39) = 0\]
So, \(y = 0\) or \(y = 39\). Since \(y = \sqrt { x^ 2 - c^ 2} \), we have:

1. \(y = 0\):
   \[\sqrt { x^ 2 - c^ 2} = 0 \implies x^ 2 - c^ 2 = 0 \implies x = \pm c\]
   Since \(c\) is positive, \(x = c\) or \(x = - c\).
2. \(y = 39\):
   \[\sqrt { x^ 2 - c^ 2} = 39 \implies x^ 2 - c^ 2 = 39^ 2 \implies x^ 2 = 39^ 2 + c^ 2 \implies x = \pm \sqrt { 39^ 2 + c^ 2} \]
   Among the options given, one of the solutions is:
   D. \(- \sqrt { c^ 2 + 39^ 2} \)

Supplemental Knowledge

Solving equations containing radicals requires isolating their radical expression and then squaring both sides of an equation to eliminate it, usually using multiple steps and careful algebraic manipulation. Furthermore, extraneous solutions might occur as you square both sides - an additional precaution must also be taken here in case any extraneous solutions surface when trying to square both sides at the same time.

From Concepts to Reality

Consider that when designing a roller coaster you must accurately calculate its hill height based on cart speed - solving for this equation accurately ensures safety and functionality for riders and the ride itself.

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