Question

5x^2+(8y-2)*x+(5y^2+2y+2)=0

Answer 1

Ответ:

$(1; -1)$

Объяснение:

$5x^{2}+8xy-2x+5y^{2}+2y+2=0;$

$5x^{2}+8xy+5y^{2}-2x+2y+2=0;$

$4x^{2}+8xy+4y^{2}+x^{2}+y^{2}-2x+2y+2=0;$

$4(x+y)^{2}+x^{2}+2xy+y^{2}-2x+2-2xy+2y=0;$

$4(x+y)^{2}+(x+y)^{2}-2(x-1)-2y(x-1)=0;$

$5(x+y)^{2}-2(x-1+y(x-1))=0;$

$5(x+y)^{2}=2(y+1)(x-1);$

$(x+y)(x+y)=0,4(y+1)(x-1);$

$\left \{ {{x+y=0,4(y+1)} \atop {x+y=x-1}} \right. \vee \left \{ {{x+y=0,4(x-1)} \atop {x+y=y+1}} \right. \Leftrightarrow \left \{ {{x+y=0,4(y+1)} \atop {y=-1}} \right. \vee \left \{ {{x+y=0,4(x-1)} \atop {x=1}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x-1=0} \atop {y=-1}} \right. \vee \left \{ {{1+y=0} \atop {x=1}} \right. \Leftrightarrow \left \{ {{x=1} \atop {y=-1}} \right. \vee \left \{ {{y=-1} \atop {x=1}} \right. ;$

$(1; -1);$

Answer 2

The equation you've provided is a quadratic equation in two variables, x and y. To solve this equation, you can use the quadratic formula. The general form of a quadratic equation is:

ax^2 + bx + c = 0

In your equation:

5x^2 + (8y - 2)x + (5y^2 + 2y + 2) = 0

You have:

a = 5 b = 8y - 2 c = 5y^2 + 2y + 2

Now, you can apply the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Plugging in the values of a, b, and c from your equation:

x = (-(8y - 2) ± √((8y - 2)² - 4(5)(5y^2 + 2y + 2))) / (2(5))

Now, you can simplify and solve for x:

x = (-8y + 2 ± √(64y² - 32y + 4 - 100y² - 40y - 40)) / 10

x = (-8y + 2 ± √(-36y² - 32y - 36)) / 10

Since the discriminant (the value inside the square root) is negative for all real values of y, this equation does not have real solutions. In other words, there are no real values of x and y that simultaneously satisfy the given equation.