¿Cuáles son los valores de x que satisfacen la desigualdad cuadrática -x^2 - 2x + 3 < 0?

Steps to Solve

1. Convert the inequality to standard form

Multiply the inequality by -1 to make the coefficient of $x^2$ positive. Remember to flip the inequality sign:

$$-x^2 - 2x + 3 < 0$$ $$x^2 + 2x - 3 > 0$$

2. Factor the quadratic expression

Factor the quadratic expression $x^2 + 2x - 3$:

$$x^2 + 2x - 3 = (x + 3)(x - 1)$$

Therefore, the inequality becomes:

$$(x + 3)(x - 1) > 0$$

3. Find the critical points (roots)

Set each factor equal to zero and solve for $x$: $$x + 3 = 0 \Rightarrow x = -3$$ $$x - 1 = 0 \Rightarrow x = 1$$

These critical points, $x = -3$ and $x = 1$, divide the number line into three intervals: $(-\infty, -3)$, $(-3, 1)$, and $(1, \infty)$.

4. Test each interval

Choose a test value within each interval and plug it into the factored inequality $(x + 3)(x - 1) > 0$ to see if the inequality holds true.

• Interval $(-\infty, -3)$: Let $x = -4$. Then $(-4 + 3)(-4 - 1) = (-1)(-5) = 5 > 0$. The inequality holds true.
• Interval $(-3, 1)$: Let $x = 0$. Then $(0 + 3)(0 - 1) = (3)(-1) = -3 > 0$. The inequality does not hold true.
• Interval $(1, \infty)$: Let $x = 2$. Then $(2 + 3)(2 - 1) = (5)(1) = 5 > 0$. The inequality holds true.

5. Write the solution

Since the inequality $x^2 + 2x - 3 > 0$ holds true for the intervals $(-\infty, -3)$ and $(1, \infty)$, the solution is:

$$x < -3 \text{ or } x > 1$$