To solve the equation $(4x - 1)^2 = -396$ using the square root property, follow these steps:

Step 1: Isolate the squared term

The equation is already in the form $(4x - 1)^2 = -396$.

Step 2: Take the square root of both sides

Taking the square root of both sides, we get: $4x - 1 = \pm \sqrt{-396}$

Step 3: Simplify the square root

Since $\sqrt{-396}$ involves a negative number under the square root, it will yield a complex number: $\sqrt{-396} = \sqrt{396} \cdot \sqrt{-1} = \sqrt{396}i$ Next, simplify $\sqrt{396}$. Since $396 = 4 \times 99$, we have: $\sqrt{396} = \sqrt{4 \times 99} = 2\sqrt{99}$ Therefore: $4x - 1 = \pm 2\sqrt{99}i$

Step 4: Solve for x

Add 1 to both sides: $4x = 1 \pm 2\sqrt{99}i$ Now, divide both sides by 4: $x = \frac{1 \pm 2\sqrt{99}i}{4}$ This simplifies to: $x = \frac{1}{4} \pm \frac{\sqrt{99}i}{2}$

Final Answer

The solutions for $x$ are: $x = \frac{1}{4} + \frac{\sqrt{99}i}{2}, \quad x = \frac{1}{4} - \frac{\sqrt{99}i}{2}$

You can enter these values in the answer boxes.

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Would you like more details or have any questions? Here are some related problems to consider:

1. Solve $(3x + 2)^2 = -100$ for $x$.
2. Solve $(5x - 7)^2 = -225$ using the square root property.
3. Find the complex roots of $x^2 + 16 = 0$.
4. Solve $(2x - 3)^2 = 49$ for $x$.
5. What is the square root of $-64$ in standard form?

Tip: Remember, the square root of a negative number always introduces an imaginary unit $i$ into the solution.