Question

(x - 4) * sqrt(x ^ 2 - x - 2) >= 0

Answer 1

Ответ:

{ - 1; 2} U [4; + ∞ ).

Объяснение:

(x - 4) • √(x² - x - 2) ≥ 0

Рассмотрим функцию

у = (x - 4) • √(x² - x - 2)

1. D(y):

x² - x - 2 ≥ 0

x² - x - 2 = 0

D = 1 + 8 = 9

x1 = (1+3)/2 = 2;

x2 = (1-3)/2 = - 1;

___+'___[-1]__-___[2]__+____ х

D(y) = (-∞;-1] U [2; !∞).

2. Нули функции:

(x - 4) • √(x² - x - 2) = 0

x - 4 = 0 или √(x² - x - 2) = 0

х = 4 ; х = 2; х = - 1.

3. ___-'___[-1]_ _ _ [2]__-___[4]___+____ х

у ≥ 0 при х є { - 1; 2} U [4; + ∞ ).

Answer 2

To solve the inequality $(x - 4) \cdot \sqrt{x^2 - x - 2} \geq 0$, you need to find the values of $x$ that satisfy this inequality.

First, find the values of $x$ that make $(x - 4)$ and $\sqrt{x^2 - x - 2}$ equal to zero.

Setting $x - 4 = 0$, you get $x = 4$.

Next, find the values of $x$ that make $\sqrt{x^2 - x - 2}$ equal to zero. Solve the quadratic equation $x^2 - x - 2 = 0$. Factoring or using the quadratic formula, you find the roots $x = -1$ and $x = 2$. However, since $\sqrt{x^2 - x - 2}$ is always non-negative, you only need to consider the interval $[-1, 2]$.

Now, test the intervals $(- \infty, -1)$, $[-1, 2]$, and $[2, \infty)$ to see where the inequality holds true. Choose test points within each interval and substitute them into the inequality $(x - 4) \cdot \sqrt{x^2 - x - 2} \geq 0$ to determine the sign of the expression in each interval.

1. For $x < -1$, choose $x = -2$: $((-2) - 4) \cdot \sqrt{(-2)^2 - (-2) - 2} = (-6) \cdot \sqrt{8} < 0$.

2. For $-1 \leq x \leq 2$, choose $x = 0$: $(0 - 4) \cdot \sqrt{0^2 - 0 - 2} = (-4) \cdot \sqrt{-2} < 0$.

3. For $x > 2$, choose $x = 3$: $(3 - 4) \cdot \sqrt{3^2 - 3 - 2} = (-1) \cdot \sqrt{4} \geq 0$.

Therefore, the inequality $(x - 4) \cdot \sqrt{x^2 - x - 2} \geq 0$ holds true for $x \in [2, \infty)$.