{x2+x-6<0 {-x2+2x+3《0

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вопрос

To solve the inequalities (x^2 + x - 6 < 0) and (-x^2 + 2x + 3 < 0), we need to find the values of (x) that satisfy both conditions.

Solve (x^2 + x - 6 < 0):
Factor the quadratic expression: (x^2 + x - 6 = (x - 2)(x + 3))
Set each factor to 0:

For (x - 2 = 0): (x = 2)
For (x + 3 = 0): (x = -3)

The critical points are (x = -3) and (x = 2). We need to determine when the quadratic expression is less than 0 (negative) between these two critical points: (-3 < x < 2).

Solve (-x^2 + 2x + 3 < 0):
Factor the quadratic expression: (-x^2 + 2x + 3 = -(x - 3)(x + 1))
Set each factor to 0:

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

The critical points are (x = -1) and (x = 3). We need to determine when the quadratic expression is less than 0 (negative) between these two critical points: (-1 < x < 3).

Now, we need to find the intersection of the intervals (-3 < x < 2) and (-1 < x < 3):
The solution to both inequalities is (-1 < x < 2), which satisfies both conditions.

Therefore, the solution to the system of inequalities is (-1 < x < 2).