Which of the following is an extraneous solution of $(45-3 x)^{\frac{1}{2}}=x-9$ ?
The given equation is \(\sqrt{45-3x} = x-9\). To solve this, we first identify the domain of the expression. For the square root to be defined in real numbers, we must have \(45-3x \geq 0\), which implies \(x \leq 15\). Furthermore, since the principal square root is non-negative, the right-hand side must satisfy \(x-9 \geq 0\), implying \(x \geq 9\). Thus, any valid solution must lie in the interval \(9 \leq x \leq 15\).
To solve the equation, we square both sides to eliminate the radical:
\[(\sqrt{45-3x})^2 = (x-9)^2\]
\[45-3x = x^2 - 18x + 81\]
Rearranging the terms to form a quadratic equation, we get:
\[x^2 - 15x + 36 = 0\]
We can factor this quadratic as:
\[(x-12)(x-3) = 0\]
The roots of this quadratic equation are \(x=12\) and \(x=3\).
Now, we verify these roots in the original equation \(\sqrt{45-3x} = x-9\):
For \(x=12\): \(\sqrt{45-3(12)} = \sqrt{45-36} = \sqrt{9} = 3\), and \(x-9 = 12-9 = 3\). Since \(3=3\), \(x=12\) is a valid solution.
For \(x=3\): \(\sqrt{45-3(3)} = \sqrt{45-9} = \sqrt{36} = 6\), and \(x-9 = 3-9 = -6\). Since \(6
eq -6\), \(x=3\) is not a solution to the original equation.
Therefore, \(x=3\) is an extraneous solution because it is a solution of the squared equation but does not satisfy the original equation. The other provided options (\(x=-12\) and \(x=-3\)) are not solutions to the squared equation and are not solutions to the original equation either.
Extraneous solutions are common in algebra when dealing with radicals, fractions, or absolute values. For instance, if an engineer is calculating the required length of a support beam based on a structural formula involving square roots, squaring both sides to solve for length might introduce values that lead to physically impossible measurements (like a negative length). Checking for extraneous solutions ensures the validity of the final design parameters, preventing structural failure or errors in complex modeling.