Question

Решите неравенство (3x−2)(x+4)>−11

Answer 1

(3x-2)(x+4)>-11
3x^2+12x-2x-8+11>0
3x^2+10x+3>0
x12=(-10+-корень(100-36))/6=(-10+-8)/6=-3 -1/3
3(x+3)(x+1/3)>0
========--3=============-1/3===========
+++++++++ ----------------------    ++++++++++
x=(-бесконечность -3)U(-1/3 +бесконечность)

Answer 2

To solve the inequality (3x - 2)(x + 4) > -11, we can follow these steps:

Step 1: Expand the product on the left side of the inequality: (3x - 2)(x + 4) = 3x^2 + 12x - 2x - 8

Simplifying further, we get: = 3x^2 + 10x - 8

So, the inequality becomes: 3x^2 + 10x - 8 > -11

Step 2: Move all the terms to one side to set the inequality to zero: 3x^2 + 10x - 8 + 11 > 0

Simplifying further, we get: 3x^2 + 10x + 3 > 0

Step 3: Solve the quadratic equation 3x^2 + 10x + 3 = 0.

To solve this equation, we can factor it or use the quadratic formula. However, let's first check if it can be factored:

3x^2 + 10x + 3 = (3x + 1)(x + 3)

The factored form shows that the equation can be factored. However, since we are dealing with an inequality, we need to find the solutions to the equation and analyze the intervals.

Setting each factor equal to zero, we have: 3x + 1 = 0 => x = -1/3 x + 3 = 0 => x = -3

Step 4: Analyze the intervals:

We have three intervals: Interval 1: x < -3 Interval 2: -3 < x < -1/3 Interval 3: x > -1/3

Step 5: Test each interval:

Choose a test point within each interval and substitute it into the inequality. If the inequality holds true for that test point, then the interval is part of the solution.

For Interval 1, let's choose x = -4: 3(-4)^2 + 10(-4) + 3 = 48 - 40 + 3 = 11 Since 11 > 0, Interval 1 is part of the solution.

For Interval 2, let's choose x = -1: 3(-1)^2 + 10(-1) + 3 = 3 - 10 + 3 = -4 Since -4 < 0, Interval 2 is not part of the solution.

For Interval 3, let's choose x = 0: 3(0)^2 + 10(0) + 3 = 3 Since 3 > 0, Interval 3 is part of the solution.

Step 6: Write the solution:

The solution to the inequality (3x - 2)(x + 4) > -11 is: x < -3 or x > -1/3

In interval notation, this can be written as: (-∞, -3) U (-1/3, ∞)