To solve the quadratic equation $x^2 - 4x + 13 = 0$, we will use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here:

• $a = 1$,
• $b = -4$,
• $c = 13$.

Step 1: Compute the discriminant

$\text{Discriminant} = b^2 - 4ac = (-4)^2 - 4(1)(13) = 16 - 52 = -36$

Since the discriminant is negative ($-36$), the solutions will involve imaginary numbers.

Step 2: Solve for $x$

$x = \frac{-(-4) \pm \sqrt{-36}}{2(1)} = \frac{4 \pm \sqrt{-36}}{2}$

The square root of $-36$ can be expressed as $6i$ (where $i = \sqrt{-1}$).

$x = \frac{4 \pm 6i}{2}$

Simplify: $x = 2 \pm 3i$

Final Answer:

The solutions are: $x = 2 + 3i \quad \text{and} \quad x = 2 - 3i$

Would you like a step-by-step breakdown or have any other questions?

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Related Questions:

1. How do you verify solutions involving complex numbers?
2. What are the applications of imaginary solutions in real-world problems?
3. How do you graph a quadratic equation with complex roots?
4. What are the conditions for discriminants leading to real, repeated, or complex solutions?
5. How does the quadratic formula change for equations with leading coefficients other than 1?

Tip: A negative discriminant always results in complex conjugate solutions!