Question

1)3х^2-11х-4<0 2)х^2-14х+40 больше 0 3)(х+3)(х+8)(х-20) больше 0 решите неравенство

Answer 1

1.

$3x^{2} - 11x < 0 \\ 3x^{2} - 11x = 0 \\ x \times (3x - 11) = 0 \\ x = 0 \\ 3x - 11 = 0 \\ 3x = 11 \\ x = 11 \div3 \\ x = \frac{11}{3}$

2.

$x^{2} - 14x + 4 > 0 \\ x {}^{2} - 14x + 4 = 0 \\ d = (- 14)^{2} - 4 \times 1 \times 4 = 180 = \sqrt{180} \\ x1 = \frac{14 + \sqrt{180} }{2} = 7 + 3 \sqrt{5} \\ x2 = \frac{14 - \sqrt{180} }{2} = 7 - 3 \sqrt{5}$

3.

$(x + 3) \times (x + 8) \times (x - 20) > 0 \\ x + 3 = 0 \\ x = - 3 \\ x + 8 = 0 \\ x = - 8 \\ x - 20 = 0 \\ x = 20$

Answer 2

To solve the given inequalities, we need to find the intervals where the expressions are either less than zero (for the first inequality) or greater than zero (for the second inequality and the third factorized inequality).

1. $3x^2 - 11x - 4 < 0$

To solve this inequality, we can find the critical points where the expression equals zero and then check the sign of the expression in the intervals between those critical points.

Step 1: Find the critical points by setting the expression equal to zero and solving for x:

$3x^2 - 11x - 4 = 0$

This quadratic equation can be factored as:

$(3x + 1)(x - 4) = 0$

Setting each factor to zero and solving for x:

$3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$

$x - 4 = 0 \Rightarrow x = 4$

Step 2: Now, let's check the sign of the expression in the intervals (-∞, -1/3), (-1/3, 4), and (4, +∞):

Test point x = 0 (in the interval (-∞, -1/3)):

$3(0)^2 - 11(0) - 4 = -4$

Test point x = -1 (in the interval (-1/3, 4)):

$3(-1)^2 - 11(-1) - 4 = 8$

Test point x = 5 (in the interval (4, +∞)):

$3(5)^2 - 11(5) - 4 = 44$

Step 3: Determine the solution:

The expression $3x^2 - 11x - 4$ is less than zero in the interval (-1/3, 4).

So, the solution to the inequality is:

$-\frac{1}{3} < x < 4$

2. $x^2 - 14x + 40 > 0$

To solve this inequality, we can again find the critical points where the expression equals zero and then check the sign of the expression in the intervals.

Step 1: Find the critical points by setting the expression equal to zero and solving for x:

$x^2 - 14x + 40 = 0$

This quadratic equation cannot be factored, so we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -14$, and $c = 40$.

$x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1}$

$x = \frac{14 \pm \sqrt{196 - 160}}{2}$

$x = \frac{14 \pm \sqrt{36}}{2}$

$x = \frac{14 \pm 6}{2}$

Now, we get two possible values for x:

$x_1 = \frac{14 + 6}{2} = 10$

$x_2 = \frac{14 - 6}{2} = 4$

Step 2: Now, let's check the sign of the expression in the intervals (-∞, 4), (4, 10), and (10, +∞):

Test point x = 3 (in the interval (-∞, 4)):

$3^2 - 14 \cdot 3 + 40 = 1 > 0$

Test point x = 6 (in the interval (4, 10)):

$6^2 - 14 \cdot 6 + 40 = 4 > 0$

Test point x = 11 (in the interval (10, +∞)):

$11^2 - 14 \cdot 11 + 40 = 1 > 0$

Step 3: Determine the solution:

The expression $x^2 - 14x + 40$ is greater than zero in the intervals (-∞, 4) and (10, +∞).

So, the solution to the inequality is:

$x < 4$ or $x > 10$

3. $(x + 3)(x + 8)(x - 20) > 0$

To solve this inequality, we can use the concept of sign changes in intervals, based on the roots of each factor.

Step 1: Find the critical points by setting each factor equal to zero and solving for x:

$x + 3 = 0 \Rightarrow x = -3$

$x + 8 = 0 \Rightarrow x = -8$

$x - 20 = 0 \Rightarrow x = 20$<