The given equation is:

$4x^2 + 1 = 0$

Step 1: Rearrange the equation

Move the constant term $1$ to the right-hand side:

$4x^2 = -1$

Step 2: Solve for $x^2$

Divide both sides by $4$:

$x^2 = -\frac{1}{4}$

Step 3: Analyze the solution

The square of a real number cannot be negative. Since $x^2 = -\frac{1}{4}$, there are no real solutions.

Step 4: Complex solutions

If we allow complex numbers, take the square root of both sides:

$x = \pm \sqrt{-\frac{1}{4}}$

Using the property of square roots, $\sqrt{-a} = i\sqrt{a}$ for $a > 0$, we have:

$x = \pm i \sqrt{\frac{1}{4}} = \pm \frac{i}{2}$

Final Answer:

The solutions are:

$x = \frac{i}{2} \quad \text{or} \quad x = -\frac{i}{2}$

Would you like a more detailed explanation or examples?

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Related Questions:

1. What is the difference between real and complex solutions?
2. How does the imaginary unit $i$ work in equations?
3. What are some applications of complex numbers in real-world problems?
4. How do you graph complex solutions on the Argand plane?
5. How can this equation change if the constant is positive instead of negative?

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Tip:

When solving quadratic equations, always check whether the discriminant ($b^2 - 4ac$) is negative to determine if the solutions are complex.