Question

Y^2+z^2-12y-8z+55>=0

Answer 1

$y^2+z^2-12y-8z+55\geq 0\\ y^2-12y+36+z^2-8z+16+3\geq 0\\ (y-6)^2+(z-4)^2+3\geq 0$

Это неравенство верно для всех действительных y,z.

Answer 2

To solve this inequality, we need to determine the region in the (Y, Z) coordinate plane where the inequality is satisfied. Let's work through the steps to find the solution:

1. Reorder the terms: y^2 - 12y + z^2 - 8z + 55 >= 0

2. Complete the square for the y and z terms: y^2 - 12y + 36 + z^2 - 8z + 16 + 55 >= 0

3. Simplify: (y - 6)^2 + (z - 4)^2 + 55 >= 0

4. Subtract 55 from both sides of the inequality: (y - 6)^2 + (z - 4)^2 >= -55

Now, we have a sum of squares on the left side of the inequality, and the right side is a negative constant. Since the square of any real number is always non-negative (greater than or equal to zero), there are no real values of y and z that satisfy the given inequality.

In conclusion, the inequality y^2 + z^2 - 12y - 8z + 55 >= 0 has no solutions in the real number system.