Solve $\sqrt{x+6}+\sqrt{x}=6$ for $x$.
The given equation is \(\sqrt{x+6} + \sqrt{x} = 6\). Before we begin, we note that for the square roots to be well-defined in the real number system, the radicands must be non-negative. This gives us the conditions \(x + 6 \ge 0\) and \(x \ge 0\). Combining these, the domain of the equation is \(x \ge 0\). Our goal is to find the value of \(x\) that satisfies this equation.
To solve the equation, we isolate one of the radical terms. Let's subtract \(\sqrt{x}\) from both sides to get:
\[\sqrt{x+6} = 6 - \sqrt{x}\]
Now, to eliminate the radical on the left side, we square both sides of the equation:
\[(\sqrt{x+6})^2 = (6 - \sqrt{x})^2\]
Expanding both sides, we use the identity \((a-b)^2 = a^2 - 2ab + b^2\):
\[x + 6 = 36 - 12\sqrt{x} + x\]
Now we simplify the equation to isolate the remaining radical term. Subtracting \(x\) from both sides gives:
\[6 = 36 - 12\sqrt{x}\]
Next, we rearrange the terms to solve for \(12\sqrt{x}\):
\[12\sqrt{x} = 36 - 6\]
\[12\sqrt{x} = 30\]
Dividing both sides by 12, we find:
\[\sqrt{x} = \frac{30}{12} = \frac{5}{2}\]
Finally, we square both sides again to solve for \(x\):
\[x = \left(\frac{5}{2}\right)^2 = \frac{25}{4}\]
We must verify our solution to ensure it is not extraneous, as squaring both sides can sometimes introduce solutions that do not satisfy the original equation. Substituting \(x = \frac{25}{4}\) back into the original equation:
\[\sqrt{\frac{25}{4} + 6} + \sqrt{\frac{25}{4}} = \sqrt{\frac{25}{4} + \frac{24}{4}} + \frac{5}{2} = \sqrt{\frac{49}{4}} + \frac{5}{2} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\]
Since this matches the original equation, the solution \(x = \frac{25}{4}\) is correct. The final answer is \(\boxed{\frac{25}{4}}\).
Radical equations often appear when analyzing physical scenarios involving square-root relationships. For instance, consider calculating the velocity of an object or determining the distance of a horizon based on altitude. If you are calculating the time it takes for an object to reach a certain state governed by square-root functions, such as the period of a pendulum or the growth rate in certain biological models, you are effectively solving equations similar to this. Understanding how to handle radicals is crucial for physics, engineering, and data modeling where relationships are non-linear.