Question

Сиситема: x+y= -4 xy= -4

Answer 1

Выразим одно через другое :
y = -4 - x

И подставим во второе выражение :
x(-4 - x) = -4
-4x - x² = -4
x² + 4x - 4 = 0

Решаю методом выделения полного квадрата:
(x + 2)² = 8

x+2 = -√8
x₁ = -2 - 2√2

x+2 = √8
x₂ = -2 + 2√2


Ответ:
(x₁ , y₁) = (-2 - 2√2 , -2 + 2√2)
(x₂ , y₂) = (-2 + 2√2 , -2 - 2√2)

Answer 2

This looks like a system of two equations with two variables. Let's solve it!

The system is:

1. $x + y = -4$
2. $xy = -4$

We can use substitution or elimination to solve. I'll use substitution here. From equation (1), we can express $x$ in terms of $y$:

$x = -4 - y$

Now substitute this expression for $x$ into equation (2):

$(-4 - y)y = -4$

Expand and simplify:

$-4y - y^2 = -4$

Rearrange into a quadratic equation:

$y^2 - 4y - 4 = 0$

Now we can solve this quadratic equation. The solutions for $y$ might be complex. Let's use the quadratic formula:

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

In our case, $a = 1$, $b = -4$, and $c = -4$. Plug these values in:

$y = \frac{4 \pm \sqrt{16 + 16}}{2}$

$y = \frac{4 \pm \sqrt{32}}{2}$

$y = \frac{4 \pm 4\sqrt{2}}{2}$

Now, simplify:

$y = 2 \pm 2\sqrt{2}$

So, the solutions for $y$ are $y = 2 + 2\sqrt{2}$ and $y = 2 - 2\sqrt{2}$.

Now that we have $y$, we can find $x$ using the expression we found earlier:

$x = -4 - y$

Substitute the values of $y$:

For $y = 2 + 2\sqrt{2}$:

$x = -4 - (2 + 2\sqrt{2}) = -6 - 2\sqrt{2}$

For $y = 2 - 2\sqrt{2}$:

$x = -4 - (2 - 2\sqrt{2}) = -2 + 2\sqrt{2}$

So, the solutions to the system are:

1. $x = -6 - 2\sqrt{2}$, $y = 2 + 2\sqrt{2}$
2. $x = -2 + 2\sqrt{2}$, $y = 2 - 2\sqrt{2}$