Question

Z^2+(5-2i)z+5(1-i)=0

Answer 1

$z^2+(5-2i)z+5(1-i)=0$

$D=(5-2i)^2-4\cdot5(1-i)=25-20i+4i^2-20+20i=25-4-20=1$

$z=\dfrac{2i-5\pm1}{2}$

$z_1=\dfrac{2i-5-1}{2} =\dfrac{2i-6}{2} =i-3$

$z_2=\dfrac{2i-5+1}{2} =\dfrac{2i-4}{2} =i-2$

Ответ: i-3; i-2

Answer 2

To solve the quadratic equation $z^2 + (5 - 2i)z + 5(1 - i) = 0$, you can use the quadratic formula:

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

In this equation, $a = 1$, $b = 5 - 2i$, and $c = 5(1 - i)$. Let's calculate the values for $a$, $b$, and $c$ first:

$a = 1$ $b = 5 - 2i$ $c = 5(1 - i) = 5 - 5i$

Now, plug these values into the quadratic formula:

$$z = \frac{-(5 - 2i) \pm \sqrt{(5 - 2i)^2 - 4(1)(5 - 5i)}}{2(1)}$$

Let's simplify the expression step by step:

1. Calculate $(5 - 2i)^2$: $(5 - 2i)^2 = 25 - 20i + 4i^2 = 25 - 20i - 4 = 21 - 20i$

2. Calculate $4(1)(5 - 5i)$: $4(1)(5 - 5i) = 20 - 20i$

Now, substitute these results back into the quadratic formula:

$$z = \frac{-(5 - 2i) \pm \sqrt{(21 - 20i) - (20 - 20i)}}{2}$$

Simplify further:

$$z = \frac{-(5 - 2i) \pm \sqrt{21 - 20i - 20 + 20i}}{2}$$

Now, the imaginary parts cancel out:

$$z = \frac{-(5 - 2i) \pm \sqrt{1}}{2}$$

Taking the square root of 1 yields two possible values:

$$z = \frac{-(5 - 2i) \pm 1}{2}$$

Now, you have two possible solutions for $z$:

1. $z_1 = \frac{-(5 - 2i) + 1}{2} = \frac{-4 - 2i}{2} = -2 - i$
2. $z_2 = \frac{-(5 - 2i) - 1}{2} = \frac{-6 + 2i}{2} = -3 + i$

So, the solutions for the quadratic equation are $z = -2 - i$ and $z = -3 + i$.