Solve for $x$.
$\sqrt{x+12}=\sqrt{x}+2$
We want to solve the equation \(\sqrt{x+12} = \sqrt{x} + 2\) for \(x\). First, we identify the domain of the equation. Since the expressions inside the square roots must be non-negative, we require \(x + 12 \geq 0\) and \(x \geq 0\). Together, this implies \(x \geq 0\). The equation involves square roots, so we will use algebraic manipulation to eliminate them and find the value of \(x\).
To eliminate the square roots, we square both sides of the given equation:
\[(\sqrt{x+12})^2 = (\sqrt{x} + 2)^2\]
The left side becomes \(x + 12\). Applying the expansion formula \((a+b)^2 = a^2 + 2ab + b^2\) to the right side, we get:
\[x + 12 = (\sqrt{x})^2 + 2(2)(\sqrt{x}) + (2)^2\]
\[x + 12 = x + 4\sqrt{x} + 4\]
Now, we simplify the equation to isolate the term containing \(\sqrt{x}\). Subtracting \(x\) from both sides, we have:
\[12 = 4\sqrt{x} + 4\]
Subtracting 4 from both sides results in:
\[8 = 4\sqrt{x}\]
Dividing both sides by 4, we obtain:
\[2 = \sqrt{x}\]
Finally, to solve for \(x\), we square both sides again:
\[(2)^2 = (\sqrt{x})^2\]
\[4 = x\]
We verify this result by substituting \(x = 4\) back into the original equation:
\(\sqrt{4+12} = \sqrt{16} = 4\)
\(\sqrt{4} + 2 = 2 + 2 = 4\)
Since both sides are equal, the solution is correct.
In physics, square root equations often arise when calculating the velocity of an object or the distance traveled under specific forces. For example, if you are determining the distance a projectile travels at a certain time, you might encounter an equation like this one. Solving for the variable allows you to predict exact movement and timing, which is essential for engineering and navigation.