Verified Answer :

  • First, add 4 to both sides of the equation to isolate the radical term: \(4\sqrt{4x} = 8\).
  • Divide both sides by 4 to obtain \(\sqrt{4x} = 2\).
  • Square both sides to eliminate the radical, resulting in \(4x = 4\).
  • Solve for \(x\) to find the final result, \(x = 1\), which is \(\boxed{1}\).

Explanation

  1. Problem Analysis and Constraints

To solve the equation \(-4 + 4\sqrt{4x} = 4\), we first identify the variable \(x\) and the constraints. For the square root \(\sqrt{4x}\) to be defined in real numbers, the expression under the radical must be non-negative, meaning \(4x \geq 0\), which simplifies to \(x \geq 0\). We are looking for the value of \(x\) that satisfies the given equation.

  1. Isolating the Radical Expression

We begin by isolating the term with the square root. Adding 4 to both sides of the equation gives:
\[4\sqrt{4x} = 4 + 4\]
\[4\sqrt{4x} = 8\]
Next, divide both sides by 4 to simplify:
\[\sqrt{4x} = \frac{8}{4}\]
\[\sqrt{4x} = 2\]

  1. Solving for x

To remove the square root, we square both sides of the equation:
\[(\sqrt{4x})^2 = 2^2\]
\[4x = 4\]
Now, solve for \(x\) by dividing both sides by 4:
\[x = \frac{4}{4}\]
\[x = 1\]

  1. Verification and Conclusion

To ensure the solution is correct, we substitute \(x = 1\) back into the original equation:
\[-4 + 4\sqrt{4(1)} = -4 + 4\sqrt{4} = -4 + 4(2) = -4 + 8 = 4\]
The result matches the right side of the original equation, confirming that \(x=1\) is indeed the correct solution. Comparing this to the given options, we find that it matches the third choice. Therefore, the final answer is \(x = 1\).

Examples

Solving equations with radicals is essential in physics and engineering, such as when calculating the velocity of an object based on its kinetic energy (\(E_k = \frac{1}{2}mv^2\), which can be rearranged to \(v = \sqrt{\frac{2E_k}{m}}\)). By isolating the radical term, you can solve for unknown variables like mass, velocity, or distance, allowing you to model various physical phenomena accurately. Mastery of these algebraic techniques provides a foundation for analyzing relationships between variables in scientific and technical fields.