Verified Answer :

  • Identify the domain constraints: \(-6 \leq x \leq 0\).
  • Square both sides to get the quadratic equation \(x^2 + 20x + 36 = 0\).
  • Factor the equation to find potential candidates: \(x = -2\) and \(x = -18\).
  • Verify the candidates against the constraints and the original equation, identifying \(x = -18\) as an extraneous solution.
  • Conclude that the only valid solution is \(\boxed{-2}\).

Explanation

  1. Analysis of the Problem and Domain Constraints

We are asked to solve the equation \(\sqrt{-8x} = x + 6\) for \(x\). First, we identify the domain of the equation. The expression inside the square root, \(-8x\), must be non-negative for the square root to be a real number, so \(-8x \geq 0\), which means \(x \leq 0\). Furthermore, the square root of a real number is non-negative, so the right-hand side \(x + 6\) must also be non-negative, meaning \(x \geq -6\). Combining these conditions, any valid solution must satisfy \(-6 \leq x \leq 0\).

  1. Squaring and Forming the Quadratic Equation

To solve for \(x\), we eliminate the square root by squaring both sides of the equation:
\[(\sqrt{-8x})^2 = (x + 6)^2\]
This simplifies to:
\[-8x = x^2 + 12x + 36\]
Next, we rearrange the equation to form a standard quadratic equation of the form \(ax^2 + bx + c = 0\):
\[x^2 + 12x + 8x + 36 = 0\]
\[x^2 + 20x + 36 = 0\]

  1. Solving the Quadratic Equation

We can solve the quadratic equation \(x^2 + 20x + 36 = 0\) by factoring. We look for two numbers that multiply to \(36\) and add up to \(20\). These numbers are \(18\) and \(2\):
\[(x + 18)(x + 2) = 0\]
Setting each factor to zero gives two candidate solutions:
\[x + 18 = 0 \implies x = -18\]
\[x + 2 = 0 \implies x = -2\]

  1. Verification and Final Result

We must now check our candidates against the domain constraints and the original equation.
First, recall the domain condition \(-6 \leq x \leq 0\).

  • For \(x = -18\), the condition is not satisfied (\(-18 < -6\)).
  • For \(x = -2\), the condition is satisfied (\(-6 \leq -2 \leq 0\)).

Next, we substitute the candidates into the original equation \(\sqrt{-8x} = x + 6\):

  • Testing \(x = -2\):

\(\sqrt{-8(-2)} = \sqrt{16} = 4\)
\(-2 + 6 = 4\)
The solution \(x = -2\) holds true.

  • Testing \(x = -18\):

\(\sqrt{-8(-18)} = \sqrt{144} = 12\)
\(-18 + 6 = -12\)
Since \(12 eq -12\), the solution \(x = -18\) is extraneous.

Therefore, the only valid solution is \(\boxed{-2}\).

Examples

This problem is an example of solving equations involving radical expressions, which appear in various physical contexts such as calculating velocities or distances where variables are under square roots. For instance, if you are determining the position \(x\) of an object whose motion is governed by a radical function related to its energy, identifying the correct root is crucial, as mathematical operations like squaring often introduce "extra" solutions that do not exist in the physical system. Knowing how to handle these extraneous roots ensures accuracy in engineering and scientific calculations.

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