Question

(x-3)(x+5)>0 как решить ?

Answer 1

Свойство произведение: +•+ = +, -•- = +

[ - знак объединения, { - пересечение

[{x-3 > 0; x > 3

[{x+5 > 0; x> -5

[

[{x-3 < 0; x < 3

[{x+5 < 0; x < -5

.

[x > 3

[x < -5

Ответ: (-∞; -5)U(3; +∞)

Answer 2

To solve the inequality $(x - 3)(x + 5) > 0$, we need to find the values of $x$ that make the expression on the left side greater than zero.

To do this, we can use the concept of intervals and test the sign of the expression for different intervals of $x$. Here's the step-by-step process:

Step 1: Find the critical points. The critical points are the values of $x$ where the expression $(x - 3)(x + 5)$ equals zero. Set each factor to zero and solve for $x$:

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

$x + 5 = 0 \Rightarrow x = -5$

Step 2: Divide the number line into intervals. We have two critical points: $x = -5$ and $x = 3$. These critical points divide the number line into three intervals: $(- \infty, -5)$, $(-5, 3)$, and $(3, +\infty)$.

Step 3: Test the sign of the expression in each interval. Choose a test point from each interval and plug it into the expression $(x - 3)(x + 5)$ to determine the sign in that interval.

For $x < -5$, let's choose $x = -6$: $(x - 3)(x + 5) = (-6 - 3)(-6 + 5) = (-9)(-1) = 9 > 0$

For $-5 < x < 3$, let's choose $x = 0$: $(x - 3)(x + 5) = (0 - 3)(0 + 5) = (-3)(5) = -15 < 0$

For $x > 3$, let's choose $x = 4$: $(x - 3)(x + 5) = (4 - 3)(4 + 5) = (1)(9) = 9 > 0$

Step 4: Determine the solution. Now, based on the sign of the expression in each interval:

• For $(- \infty, -5)$, the expression is positive (+).
• For $(-5, 3)$, the expression is negative (-).
• For $(3, +\infty)$, the expression is positive (+).

The inequality $(x - 3)(x + 5) > 0$ is satisfied when the expression is positive (+). So, the solution is the union of the intervals where the expression is positive: $(- \infty, -5)$ and $(3, +\infty)$.

In interval notation, the solution is $x \in (-\infty, -5) \cup (3, +\infty)$. This means any real number less than -5 or greater than 3 will satisfy the inequality.