Solve for $x$.
$\sqrt{x+12}=\sqrt{x}+2$
The given equation is \(\sqrt{x+12} = \sqrt{x} + 2\). We first determine the domain of validity for the expression. For the square roots to be real-valued, we require \(x \geq 0\) and \(x + 12 \geq 0\). Thus, the domain is \(x \geq 0\). We aim to isolate the variable \(x\) and verify if the result satisfies the original equation.
To eliminate the radicals, we begin by squaring both sides of the equation:
\[(\sqrt{x+12})^2 = (\sqrt{x} + 2)^2\]
This simplifies the left-hand side to \(x+12\). Expanding the right-hand side using the identity \((a+b)^2 = a^2 + 2ab + b^2\), where \(a = \sqrt{x}\) and \(b = 2\), we obtain:
\[x + 12 = x + 4\sqrt{x} + 4\]
Next, we simplify the equation to isolate the term containing the square root:
\[x + 12 = x + 4\sqrt{x} + 4\]
Subtract \(x\) from both sides:
\[12 = 4\sqrt{x} + 4\]
Subtract 4 from both sides:
\[8 = 4\sqrt{x}\]
Divide by 4:
\[2 = \sqrt{x}\]
Finally, we solve for \(x\) by squaring both sides once more:
\[(2)^2 = (\sqrt{x})^2\]
\[4 = x\]
To ensure the solution is valid, we substitute \(x = 4\) back into the original equation \(\sqrt{x+12} = \sqrt{x} + 2\):
\(\sqrt{4+12} = \sqrt{16} = 4\)
\(\sqrt{4} + 2 = 2 + 2 = 4\)
Since both sides are equal, the solution is correct. The result is \(x = 4\).
Square root equations are frequently used in physics and engineering, for example when calculating the distance an object has fallen or determining velocity in kinematics problems. If a situation involves the square root of time or position, squaring the equation to eliminate radicals is a fundamental skill that allows for the straightforward determination of the unknown variable, much like in this problem where we determined the value of x.