Question

1) x2 + 4x + 13 = 0​

Answer 1

Ответ:

решения нет, так как дискриминант меньше 0

Answer 2

To solve the quadratic equation $x^2 + 4x + 13 = 0$, you can use the quadratic formula:

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

In this equation, $a = 1$, $b = 4$, and $c = 13$. Plug these values into the formula:

$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(13)}}{2(1)}$

Now, calculate the values under the square root:

$x = \frac{-4 \pm \sqrt{16 - 52}}{2}$ $x = \frac{-4 \pm \sqrt{-36}}{2}$

Since the expression under the square root ($-36$) is negative, the quadratic equation has no real solutions. Instead, it has two complex solutions because you're dealing with the square root of a negative number. The complex solutions are of the form $x = \frac{-b}{2a} \pm \frac{\sqrt{-36}}{2a}i$:

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

Simplify:

$x = -2 \pm 3i$

So, the solutions to the equation $x^2 + 4x + 13 = 0$ are $x = -2 + 3i$ and $x = -2 - 3i$, where $i$ represents the imaginary unit.