Question

Розв'язати нерівності A) (x-1)(x+5)>0 B) -x²+2x-7<0

Answer 1

Объяснение:

А) правильно

В) не правильно

Answer 2

A) (x-1)(x+5) > 0

To solve this inequality, we can use the concept of sign charts. We need to find the intervals where the expression (x-1)(x+5) is greater than 0, meaning positive.

1. First, find the critical points where the expression equals 0: (x-1)(x+5) = 0 Set each factor equal to 0 and solve for x: x - 1 = 0 => x = 1 x + 5 = 0 => x = -5

2. Now, create a sign chart using these critical points: Place -5 and 1 on the number line:

   (-∞)-----(-5)-----[1]-----(∞)

3. Pick test points in each of the three intervals:

   • Test x = -6 in the interval (-∞, -5).
   • Test x = 0 in the interval (-5, 1).
   • Test x = 2 in the interval (1, ∞).

4. Evaluate the expression (x-1)(x+5) for each test point:

   • For x = -6: (-6 - 1)(-6 + 5) = (-7)(-1) = 7 (positive)
   • For x = 0: (0 - 1)(0 + 5) = (-1)(5) = -5 (negative)
   • For x = 2: (2 - 1)(2 + 5) = (1)(7) = 7 (positive)

5. Determine the sign of the expression in each interval:

   • (-∞, -5): Positive
   • (-5, 1): Negative
   • (1, ∞): Positive

6. The original inequality is satisfied when the expression is greater than 0, so the solution is: x in (-∞, -5) U (1, ∞)

B) -x² + 2x - 7 < 0

This is a quadratic inequality. To solve it, we can use a different approach:

1. First, we can multiply the entire inequality by -1 to make it easier to work with: x² - 2x + 7 > 0

2. Now, let's solve for the roots of the related equation: x² - 2x + 7 = 0

The discriminant, Δ = b² - 4ac, is less than 0, which means there are no real roots. Therefore, the quadratic is always positive (it opens upward), and there are no x-values that make it equal to zero.

3. Since the quadratic is always positive, the inequality x² - 2x + 7 > 0 is true for all real values of x.

So, the solution to the inequality is: x ∈ (-∞, ∞)