What is the solution of $\sqrt{x+12}=x$ ?
We are given the radical equation \(\sqrt{x+12} = x\). To solve this, we must consider the domain of the functions involved. Since the square root function \(\sqrt{u}\) returns non-negative values, the right side of the equation must be non-negative. Thus, we must have \(x \geq 0\). Additionally, the expression inside the square root must be non-negative: \(x+12 \geq 0\), which implies \(x \geq -12\). Combining these conditions, any valid solution must satisfy \(x \geq 0\).
To eliminate the radical, we square both sides of the equation:
\[(\sqrt{x+12})^2 = x^2\]
\[x + 12 = x^2\]
Rearranging this into a standard quadratic equation format gives:
\[x^2 - x - 12 = 0\]
We factor the quadratic equation \(x^2 - x - 12 = 0\). We look for two numbers that multiply to \(-12\) and add to \(-1\). These numbers are \(-4\) and \(3\). Thus, the equation factors as:
\[(x - 4)(x + 3) = 0\]
Setting each factor to zero, we find the potential roots:
\[x - 4 = 0 \Rightarrow x = 4\]
\[x + 3 = 0 \Rightarrow x = -3\]
We must verify our potential solutions against the original equation and our constraint \(x \geq 0\).
For \(x = 4\): The constraint \(4 \geq 0\) is satisfied. Plugging into the original equation, we get \(\sqrt{4+12} = \sqrt{16} = 4\), which is correct.
For \(x = -3\): The constraint \(-3 \geq 0\) is not satisfied. Plugging into the original equation, we get \(\sqrt{-3+12} = \sqrt{9} = 3\), which does not equal \(-3\). Thus, \(x = -3\) is an extraneous root.
Therefore, the only valid solution is \(x = 4\).
Radical equations like \(\sqrt{x+12} = x\) are commonly used in physics and engineering, such as determining the speed of an object or distance based on time constraints. For example, if the distance traveled by an object is modeled by a square root function related to time, finding the correct time \(x\) when the object reaches a specific distance is vital. Ensuring the domain of the variable is respected avoids physically impossible results, like a negative time or distance, just as we identified the extraneous root in our equation.