Question

X^2-2x-3 больше или равно 0 решить надо методом интервала

Answer 1

X^2 - 2x - 3 ≥ 0 
x^2 - 2x - 3 = 0  -  Приравняли к нулю.
D= b^2 - 4ac = ( - 2) ^2 - 4 * 1 *( -3) = 4+12=16
√D=4
x1 = 2 + 4 / 2 = 3
x2 = 2 - 4 / 2 = - 1 
Рисунок во вложении.
Ответ: x ∈  ( - ∞ ; - 1 ]  U [ 3 ; + ∞ ) 

Answer 2

Solving the Equation Using the Interval Method

To solve the equation x^2 - 2x - 3 ≥ 0 using the interval method, we need to find the intervals where the equation is greater than or equal to zero.

Let's break down the process step by step:

1. Find the critical points of the equation by setting it equal to zero and solving for x: x^2 - 2x - 3 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring the equation, we get: (x - 3)(x + 1) = 0

Setting each factor equal to zero, we find the critical points: x - 3 = 0 or x + 1 = 0 x = 3 or x = -1

Therefore, the critical points are x = 3 and x = -1.

2. Determine the intervals using the critical points: We can divide the number line into three intervals based on the critical points: (-∞, -1), (-1, 3), and (3, +∞).

3. Test a point within each interval to determine the sign of the equation: - For the interval (-∞, -1), we can choose x = -2 as a test point. Substituting this value into the equation, we get: (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5

Since the result is positive, the equation is greater than zero in this interval.

- For the interval (-1, 3), we can choose x = 0 as a test point. Substituting this value into the equation, we get: 0^2 - 2(0) - 3 = -3

Since the result is negative, the equation is less than zero in this interval.

- For the interval (3, +∞), we can choose x = 4 as a test point. Substituting this value into the equation, we get: 4^2 - 2(4) - 3 = 16 - 8 - 3 = 5

Since the result is positive, the equation is greater than zero in this interval.

4. Determine the solution: Based on the signs of the equation in each interval, we can conclude that the equation x^2 - 2x - 3 ≥ 0 is true for the intervals (-∞, -1) and (3, +∞).

Therefore, the solution to the equation using the interval method is: x ≤ -1 or x ≥ 3.

Please note that the solution is inclusive of the critical points (-1 and 3) because the equation is greater than or equal to zero.

I hope this helps! Let me know if you have any further questions.