Question

(X-2)^2 больше либо равно x(x-3)+8

Answer 1

(x-2)^2 ≥x(x-3)+8
x²-4x+4≥x²-3x+8
-3x+4x≤4-8
x≤-4
ответ x∈(-∞;-4]

Answer 2

(x-2) $^{2}$ ≥x(x-3)+8
-3x+4x≤4-8
x≤-4
Ответ: от (- бесконечности до - 4]

Answer 3

To determine whether (X-2)^2 is greater than or equal to x(x-3) + 8, we need to compare the two expressions.

Let's first expand (X-2)^2: (X-2)^2 = (X-2)(X-2) = X^2 - 4X + 4

Now, the given expression is x(x-3) + 8: x(x-3) + 8 = x^2 - 3x + 8

Now, we need to compare the two expressions: (X-2)^2 >= x(x-3) + 8

Substitute the expanded form of (X-2)^2 and x(x-3) + 8: X^2 - 4X + 4 >= x^2 - 3x + 8

Now, let's simplify the inequality by moving all terms to one side: X^2 - x^2 - 4X + 3x + 4 - 8 >= 0

Combine like terms: -X^2 - X - 4 >= 0

Now, we have a quadratic inequality. To solve it, we need to find the critical points where the expression equals zero. Then, we'll check the sign of the expression in the intervals between these critical points.

To find the critical points, set the expression equal to zero and solve for X: -X^2 - X - 4 = 0

Multiply by -1 to simplify: X^2 + X + 4 = 0

The quadratic equation doesn't have real solutions since its discriminant (b^2 - 4ac) is negative. Therefore, there are no critical points to check.

Now, let's analyze the coefficient of the X^2 term, which is -1 (negative). Since the coefficient is negative, the parabola opens downwards. Thus, the expression will be negative for all values of X.

Conclusion: (X-2)^2 is never greater than or equal to x(x-3) + 8 for any real value of X.