Question

0;9) (3x + 4) (2x - 5) > 0;11) (3x - 4) (2x + 7) > 0​

Answer 1

Ответ:

решение смотри на фотографии

Answer 2

To solve inequalities involving quadratic expressions, you can use a sign chart or test points to determine the intervals where the expressions are greater than zero.

Let's analyze both inequalities:

1. (3x + 4)(2x - 5) > 0

To solve this inequality, you can find the critical points where the expression equals zero, which are the values of x that make each factor zero. Then, you can use test intervals to determine where the expression is positive.

a) Set each factor equal to zero and solve for x: 3x + 4 = 0 => 3x = -4 => x = -4/3 2x - 5 = 0 => 2x = 5 => x = 5/2

So, you have two critical points, x = -4/3 and x = 5/2.

b) Now, create a sign chart for the expression: |---|---|---|---|---| -∞ -4/3 5/2 ∞

c) Choose a test point in each interval and plug it into the expression to determine its sign:

• For x < -4/3, you can choose x = -2. (3(-2) + 4)(2(-2) - 5) = (-6 + 4)(-4 - 5) = (-2)(-9) = 18 Since this is positive, the expression is > 0 in this interval.

• For -4/3 < x < 5/2, you can choose x = 0. (3(0) + 4)(2(0) - 5) = (4)(-5) = -20 Since this is negative, the expression is < 0 in this interval.

• For x > 5/2, you can choose x = 3. (3(3) + 4)(2(3) - 5) = (9 + 4)(6 - 5) = (13)(1) = 13 Since this is positive, the expression is > 0 in this interval.

Now, combine the results:

(3x + 4)(2x - 5) > 0 for x < -4/3 and x > 5/2.

2. (3x - 4)(2x + 7) > 0

Repeat the same process for this inequality:

a) Set each factor equal to zero and solve for x: 3x - 4 = 0 => 3x = 4 => x = 4/3 2x + 7 = 0 => 2x = -7 => x = -7/2

You have two critical points, x = 4/3 and x = -7/2.

b) Create a sign chart: |---|---|---|---|---| -∞ -7/2 4/3 ∞

c) Test the intervals:

• For x < -7/2, choose x = -3. (3(-3) - 4)(2(-3) + 7) = (-9 - 4)(-6 + 7) = (-13)(1) = -13 The expression is < 0 in this interval.

• For -7/2 < x < 4/3, choose x = 1. (3(1) - 4)(2(1) + 7) = (3 - 4)(2 + 7) = (-1)(9) = -9 The expression is < 0 in this interval.

• For x > 4/3, choose x = 2. (3(2) - 4)(2(2) + 7) = (6 - 4)(4 + 7) = (2)(11) = 22 The expression is > 0 in this interval.

Combine the results:

(3x - 4)(2x + 7) > 0 for x > 4/3.

So, the solution to the inequalities is:

1. (3x + 4)(2x - 5) > 0 for x < -4/3 and x > 5/2.
2. (3x - 4)(2x + 7) > 0 for x > 4/3.