Question

5.43. 1) (x² - 4)(2x - 1) < 0; 3) (x - 1)(x + 2)(3x - 1)>0; 2) (9 – x°)(6 – 5x)>0; 4) (2x - 5)(x +0,5)) (3x + 7) <0.

Answer 1

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Answer 2

To solve these inequalities, we need to find the intervals of x that satisfy each inequality. Let's solve them one by one:

1. (x² - 4)(2x - 1) < 0

First, we find the critical points by setting each factor equal to zero: x² - 4 = 0 (x - 2)(x + 2) = 0 x = 2 or x = -2

Next, we create a number line and test the intervals between and outside the critical points:

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   -3       -2       2       3
   |--------|--------|--------|

We pick test points in each interval and check if the expression is positive or negative:

In the interval (-∞, -2): Pick x = -3: (-3² - 4)(2(-3) - 1) = (9 - 4)(-6 - 1) = 5(-7) = -35 < 0

In the interval (-2, 2): Pick x = 0: (0² - 4)(2(0) - 1) = (-4)(-1) = 4 > 0

In the interval (2, ∞): Pick x = 3: (3² - 4)(2(3) - 1) = (9 - 4)(6 - 1) = 5(5) = 25 > 0

Therefore, the solution to the inequality is x ∈ (-∞, -2) ∪ (2, ∞).

2. (9 - x°)(6 - 5x) > 0

Simplify the expression: (9 - x)(6 - 5x) > 0

Set each factor equal to zero to find the critical points: 9 - x = 0 x = 9

6 - 5x = 0 x = 6/5

Create a number line and test the intervals:

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   6/5      9
   |--------|--------|

In the interval (-∞, 6/5): Pick x = 0: (9 - 0)(6 - 5(0)) = 9(6) = 54 > 0

In the interval (6/5, 9): Pick x = 7: (9 - 7)(6 - 5(7)) = 2(-29) = -58 < 0

In the interval (9, ∞): Pick x = 10: (9 - 10)(6 - 5(10)) = (-1)(-44) = 44 > 0

The solution to the inequality is x ∈ (-∞, 6/5) ∪ (9, ∞).

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

Set each factor equal to zero: x - 1 = 0 x = 1

x + 2 = 0 x = -2

3x - 1 = 0 x = 1/3

Create a number line and test the intervals:

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   -3       -2       1/3       1       2
   |--------|--------|--------|--------|

In the interval (-∞, -2): Pick x = -3: (-3 - 1)(-3 + 2)(3(-3) - 1) = (-4)(-1)(-10) = 40 > 0

In the interval (-2,