Which of the following is an extraneous solution of $\sqrt{-3 x-2}=x+2$?
To find the extraneous solution for the given radical equation, we first perform algebraic manipulations to find potential solutions. The equation is \(\sqrt{-3x - 2} = x + 2\). Squaring both sides of the equation, we get:
\[(\sqrt{-3x - 2})^2 = (x + 2)^2\]
\[-3x - 2 = x^2 + 4x + 4\]
Rearranging this into a standard quadratic form \(Ax^2 + Bx + C = 0\):
\[x^2 + 4x + 3x + 4 + 2 = 0\]
\[x^2 + 7x + 6 = 0\]
Factoring the quadratic expression:
\[(x + 6)(x + 1) = 0\]
Thus, the potential solutions are \(x = -6\) and \(x = -1\).
To identify which solution is extraneous, we substitute each potential value back into the original equation \(\sqrt{-3x - 2} = x + 2\).
Case 1: \(x = -6\)
Left side (LHS): \(\sqrt{-3(-6) - 2} = \sqrt{18 - 2} = \sqrt{16} = 4\)
Right side (RHS): \(-6 + 2 = -4\)
Since \(4
eq -4\), \(x = -6\) is an extraneous solution.
Case 2: \(x = -1\)
Left side (LHS): \(\sqrt{-3(-1) - 2} = \sqrt{3 - 2} = \sqrt{1} = 1\)
Right side (RHS): \(-1 + 2 = 1\)
Since \(1 = 1\), \(x = -1\) is a valid solution.
Based on our check, \(x = -6\) does not satisfy the original equation, while \(x = -1\) does. Therefore, the extraneous solution of the given equation is \(\boxed{-6}\).
Extraneous solutions occur when solving equations that involve non-invertible operations, such as squaring, which can introduce "false" roots. Think of this like balancing a scale: if you square both sides of an equation, you're essentially changing the properties of the balance. For example, if you have a weight \(y = 4\) on one side, it is clearly positive. However, if you square it, you get \(y^2 = 16\), and the equation \(\sqrt{16} = y\) could be satisfied by both \(4\) and \(-4\). In practical applications like engineering or physics, verifying solutions is crucial to ensure that the results are physically meaningful and don't lead to errors in system performance. Always checking your work helps avoid these phantom solutions!