Question

(x^2 +x+ 2)(х – 4) < 0;​

Answer 1

Пошаговое объяснение:

вот так надо решать...

Answer 2

To solve the inequality $(x^2 + x + 2)(x - 4) < 0$, we can use a method called the sign chart.

Step 1: Find the critical points. To determine the critical points, we set each factor equal to zero and solve for $x$: $x^2 + x + 2 = 0$ has no real solutions because the discriminant is negative. $x - 4 = 0$ gives $x = 4$.

Step 2: Create a sign chart. We consider three intervals based on the critical points: $(-\infty, 4)$, $(4, \infty)$, and $x = 4$ (we include this point since the inequality is not strict).

Step 3: Determine the sign of each factor in each interval. For the factor $x^2 + x + 2$, since the coefficient of the $x^2$ term is positive, the parabola opens upward, indicating that the expression is positive for all real values of $x$.

For the factor $x - 4$, we can use test points to determine the sign. Choosing a test point from each interval:

• For $x = 0$, $(0 - 4) = -4$ which is negative.
• For $x = 5$, $(5 - 4) = 1$ which is positive.

Step 4: Determine the sign of the whole expression. To determine the sign of the whole expression, we multiply the signs of each factor together:

$(x^2 + x + 2)(x - 4) < 0$

• In the interval $(-\infty, 4)$, we have (+)(-) = (-), indicating that the expression is negative in this interval.
• At $x = 4$, the expression equals 0.
• In the interval $(4, \infty)$, we have (+)(+) = (+), indicating that the expression is positive in this interval.

Step 5: Write the solution. From the sign chart, we see that the inequality $(x^2 + x + 2)(x - 4) < 0$ holds true for $x$ in the interval $(-\infty, 4)$.

Therefore, the solution to the inequality is $x \in (-\infty, 4)$.