Question

Z^2+14z+33=0 z1=? z2=? ^ - степень

Answer 1

Ответ:

${z}^{2} + 14z + 33 = 0 \\ d = {b}^{2} - 4ac = {14}^{2} - 4 \times 1 \times 33 = 64 \\ \sqrt{d} = \sqrt{64} = 8 \\ d > 0$

Если дискиминант больше 0,значит существует 2 корня

$z1 = - 14 + 8 = 6 \\ z2 = - 14 - 8 = - 22$

Answer 2

To find the solutions for the quadratic equation $z^2 + 14z + 33 = 0$, we can use the quadratic formula:

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

In this equation, $a = 1$, $b = 14$, and $c = 33$.

Let's plug in these values and solve for $z_1$ and $z_2$:

$z = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 1 \cdot 33}}{2 \cdot 1}$

$z = \frac{-14 \pm \sqrt{196 - 132}}{2}$

$z = \frac{-14 \pm \sqrt{64}}{2}$

$z = \frac{-14 \pm 8}{2}$

Now we have two possible solutions:

$z_1 = \frac{-14 + 8}{2} = -3$ $z_2 = \frac{-14 - 8}{2} = -11$

So, the solutions to the quadratic equation $z^2 + 14z + 33 = 0$ are $z_1 = -3$ and $z_2 = -11$.