Question

1) -x² - 2x +3>0 2) x² ≤ 25

Answer 1

Ответ:

1) Решим неравенство -x² - 2x +3>0:

Сначала найдем корни квадратного трехчлена -x² - 2x +3 = 0:

D = (-2)² - 4*(-1)*3 = 16 > 0, корней два:

x1,2 = (-(-2) ± √16)/(-2) = 1 ± 2

x1 = -1, x2 = 3

Построим таблицу знаков:

x | -∞ |-1 | 3 | +∞

---|---|---|---|---

-x²-2x+3>0| + | - | + | +

Ответ: (-1, 3)

2) Решим неравенство x² ≤ 25:

x² - 25 ≤ 0

(x - 5)(x + 5) ≤ 0

Построим таблицу знаков:

x | -∞ |-5 | 5 | +∞

---|---|---|---|---

(x-5)(x+5)| - | - | + | +

Ответ: (-∞, -5] ∪ [5, +∞)

Answer 2

To solve these inequalities, you can break them down one by one.

1. -x² - 2x + 3 > 0:

First, let's find the roots of the quadratic equation -x² - 2x + 3 = 0 by factoring or using the quadratic formula:

x² + 2x - 3 = 0

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

So, the roots are x = -3 and x = 1.

Now, we need to determine the sign of the quadratic expression in different intervals:

Interval 1: x < -3 Choose a test point, say x = -4, and plug it into the expression: -x² - 2x + 3 -(-4)² - 2(-4) + 3 = -16 + 8 + 3 = -5 Since -5 < 0, the expression is negative in this interval.

Interval 2: -3 < x < 1 Choose a test point, say x = 0, and plug it into the expression: -x² - 2x + 3 -(0)² - 2(0) + 3 = 3 Since 3 > 0, the expression is positive in this interval.

Interval 3: x > 1 Choose a test point, say x = 2, and plug it into the expression: -x² - 2x + 3 -(2)² - 2(2) + 3 = -4 - 4 + 3 = -5 Since -5 < 0, the expression is negative in this interval.

Now, to solve the inequality -x² - 2x + 3 > 0, we look for the intervals where the expression is greater than zero:

Interval 2: -3 < x < 1

So, the solution to the inequality is:

-3 < x < 1

2. x² ≤ 25:

To solve this inequality, you can take the square root of both sides, but remember to consider both the positive and negative roots:

√(x²) ≤ √(25)

|x| ≤ 5

This means that x can be any real number such that its absolute value is less than or equal to 5. In interval notation:

-5 ≤ x ≤ 5

So, the solution to the inequality x² ≤ 25 is:

-5 ≤ x ≤ 5