Question

2x^2-3xy+y^2=0 y^2-x^2=12

Answer 1

Ответ:

Ответ закреплён в фотографии

Answer 2

To solve this system of equations, you can use the method of substitution. First, isolate one of the variables in one of the equations and then substitute it into the other equation. Let's solve it step by step:

Given equations:

1. $2x^2 - 3xy + y^2 = 0$
2. $y^2 - x^2 = 12$

First, we can solve the second equation for $y^2$:

$y^2 = x^2 + 12$

Now, we can substitute this expression for $y^2$ into the first equation:

$2x^2 - 3xy + (x^2 + 12) = 0$

Now, simplify the equation:

$3x^2 - 3xy + 12 = 0$

Next, we can factor out a common factor of 3:

$3(x^2 - xy + 4) = 0$

Now, we have a quadratic equation in terms of $x$:

$x^2 - xy + 4 = 0$

This quadratic equation can be solved using the quadratic formula:

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

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

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

Now, calculate the discriminant:

$1 - 4(1)(4) = 1 - 16 = -15$

Since the discriminant is negative, there are no real solutions for $x$. However, there can be complex solutions. Let's proceed:

$x = \frac{1 \pm \sqrt{-15}}{2}$

The square root of -15 is an imaginary number, so the solutions for $x$ will be complex numbers:

$x = \frac{1 \pm i\sqrt{15}}{2}$

Now, let's find the corresponding values of $y$ using the equation $y^2 = x^2 + 12$:

For $x = \frac{1 + i\sqrt{15}}{2}$:

$y^2 = \left(\frac{1 + i\sqrt{15}}{2}\right)^2 + 12$

You can simplify this expression to find the corresponding complex values of $y$.

For $x = \frac{1 - i\sqrt{15}}{2}$, you can follow the same process to find the corresponding complex values of $y$.

So, the solutions to the system of equations are complex pairs of $(x, y)$ values.