Question

A) a² + 2a + 2 > 0;​

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

To determine the values of "a" for which the inequality $a^2 + 2a + 2 > 0$ is true, you can use the quadratic inequality approach.

First, let's find the roots of the quadratic equation $a^2 + 2a + 2 = 0$. The roots can be found using the quadratic formula:

$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = 2$, and $c = 2$, so:

$a = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$ $a = \frac{-2 \pm \sqrt{4 - 8}}{2}$ $a = \frac{-2 \pm \sqrt{-4}}{2}$

Since the discriminant $\sqrt{-4}$ is an imaginary number ($2i$), there are no real roots for this equation. This means that the quadratic $a^2 + 2a + 2$ does not intersect the x-axis, and it is always positive.

Therefore, for all real values of "a," $a^2 + 2a + 2$ is greater than zero, and the solution to the inequality $a^2 + 2a + 2 > 0$ is:

$-\infty < a < +\infty$

In other words, the inequality is true for all real values of "a."