Question

4x²-225≤0 Решите пожайлутса :

Answer 1

4x²≤225
x²≤225/4
x≤5/2

Answer 2

Solving the Inequality 4x² - 225 ≤ 0

To solve the inequality 4x² - 225 ≤ 0, we need to find the values of x that satisfy the inequality. Let's break down the steps to solve it:

Step 1: Factor the quadratic expression The expression 4x² - 225 can be factored as the difference of squares: (2x - 15)(2x + 15).

Step 2: Set each factor equal to zero and solve for x Setting (2x - 15) equal to zero gives us: 2x - 15 = 0 Solving for x, we get: 2x = 15 x = 15/2 x = 7.5

Setting (2x + 15) equal to zero gives us: 2x + 15 = 0 Solving for x, we get: 2x = -15 x = -15/2 x = -7.5

Step 3: Determine the intervals where the inequality is satisfied Now that we have the critical points x = 7.5 and x = -7.5, we can determine the intervals where the inequality is satisfied. We can do this by testing values within each interval.

Let's consider the interval x < -7.5: If we choose a value within this interval, such as x = -10, we can substitute it into the inequality: 4(-10)² - 225 ≤ 0 400 - 225 ≤ 0 175 ≤ 0

Since 175 is not less than or equal to 0, this interval does not satisfy the inequality.

Next, let's consider the interval -7.5 ≤ x ≤ 7.5: If we choose a value within this interval, such as x = 0, we can substitute it into the inequality: 4(0)² - 225 ≤ 0 0 - 225 ≤ 0 -225 ≤ 0

Since -225 is less than or equal to 0, this interval satisfies the inequality.

Finally, let's consider the interval x > 7.5: If we choose a value within this interval, such as x = 10, we can substitute it into the inequality: 4(10)² - 225 ≤ 0 400 - 225 ≤ 0 175 ≤ 0

Since 175 is not less than or equal to 0, this interval does not satisfy the inequality.

Solution:

The solution to the inequality 4x² - 225 ≤ 0 is the interval -7.5 ≤ x ≤ 7.5. This means that any value of x within this interval, including the endpoints, will satisfy the inequality.

Please note that the critical points and intervals were determined by factoring the quadratic expression and testing values within each interval.