Question

4x^2+y^2-4x-14y+50=0

Answer 1

Ответ:

решение представлено на фото

Answer 2

(2x-1)^2+(y-7)^2=0
x=0,5 y=7

Answer 3

The equation you provided is a quadratic equation in two variables, x and y. To analyze it further, let's rearrange the equation:

4x^2 - 4x + y^2 - 14y + 50 = 0

To determine the shape of the equation, we can group the x-terms together and the y-terms together:

(4x^2 - 4x) + (y^2 - 14y) + 50 = 0

Now, complete the square for both the x-terms and y-terms:

4(x^2 - x) + (y^2 - 14y) + 50 = 0

To complete the square for the x-terms, we take half of the coefficient of x (-1), square it (1), and add it inside the parentheses. However, we must also subtract the same value outside the parentheses to maintain equality:

4(x^2 - x + 1/4 - 1/4) + (y^2 - 14y) + 50 = 0

Simplifying further:

4((x - 1/2)^2 - 1/4) + (y^2 - 14y) + 50 = 0

Similarly, for the y-terms:

4((x - 1/2)^2 - 1/4) + (y^2 - 14y + 49 - 49) + 50 = 0

Simplifying:

4(x - 1/2)^2 + (y - 7)^2 - 49 + 50 = 0

4(x - 1/2)^2 + (y - 7)^2 = -1

Since the sum of squares cannot be negative, there are no real solutions for this equation. The graph of this equation would be an empty set or no graph at all. It is likely that there was a typo or error in the equation you provided.