Question

Решите неравенство: х²-х-12≤0

Answer 1

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Answer 2

Х²-х-12≤0
Д=1+48=49=7²
х=(1±7)/2
х1=4;х2=-3
х²-х-12=(х-4)(х+3)≤0

по методу интервалов
____+___-3____-____4__+_____
х€[-3;4]

Answer 3

To solve the inequality $x^2 - x - 12 \leq 0$, we can use the method of factoring. First, let's factor the left-hand side expression:

$x^2 - x - 12 = (x - 4)(x + 3)$

Now, we need to find the critical points where the expression equals zero:

$x - 4 = 0 \quad \text{or} \quad x + 3 = 0$

Solving for $x$ in each case, we get:

1. $x - 4 = 0 \Rightarrow x = 4$
2. $x + 3 = 0 \Rightarrow x = -3$

Now, we have three intervals on the number line separated by these critical points: $-\infty < -3 < 4 < \infty$. We can test a point in each interval to determine the sign of $x^2 - x - 12$. Let's pick $x = 0$ for the first interval, $x = -2$ for the second interval, and $x = 5$ for the third interval:

1. For $x = 0$: $x^2 - x - 12 = (0 - 4)(0 + 3) = -12 < 0$
2. For $x = -2$: $x^2 - x - 12 = (-2 - 4)(-2 + 3) = 6 > 0$
3. For $x = 5$: $x^2 - x - 12 = (5 - 4)(5 + 3) = 8 > 0$

Now, we can summarize the results for each interval:

1. $-\infty < x < -3$: $x^2 - x - 12 < 0$
2. $-3 \leq x \leq 4$: $x^2 - x - 12 \geq 0$
3. $4 < x < \infty$: $x^2 - x - 12 > 0$

The inequality $x^2 - x - 12 \leq 0$ is satisfied for $x$ in the interval $-\infty < x \leq -3$ and $4 \leq x < \infty$. So, the solution to the inequality is:

$-\infty < x \leq -3$ or $4 \leq x < \infty$