Question

A) x^2 - 6x + 8 >0;б) x^2 + 6x + 8 < 0;B) – x^2 – 2x + 15 <0​

Answer 1

Ответ:

Это правильно. Можно лучший ответ

Answer 2

вроде так......... оценку пж

Answer 3

To solve the inequalities, let's break them down one by one:

A) $x^2 - 6x + 8 > 0$

We can start by factoring the left side:

$x^2 - 6x + 8 = (x - 2)(x - 4)$

Now, we need to find the critical points where $x^2 - 6x + 8 = 0$, which are $x = 2$ and $x = 4$.

Next, we'll test the intervals created by these critical points:

1. $x < 2$: Pick a test point, for example, $x = 0$. Plug it into $(x - 2)(x - 4)$, which gives $(0 - 2)(0 - 4) = 8 > 0$. This interval satisfies the inequality.
2. $2 < x < 4$: Use $x = 3$ as a test point. $(3 - 2)(3 - 4) = -1 < 0$. This interval does not satisfy the inequality.
3. $x > 4$: Use $x = 5$ as a test point. $(5 - 2)(5 - 4) = 3 > 0$. This interval satisfies the inequality.

So, the solution for A is $x < 2$ or $x > 4$.

B) $-x^2 - 2x + 15 < 0$

Let's multiply both sides by -1 to simplify:

$x^2 + 2x - 15 > 0$

We can factor the left side:

$x^2 + 2x - 15 = (x + 5)(x - 3)$

Finding the critical points where $x^2 + 2x - 15 = 0$ gives $x = -5$ and $x = 3$.

Testing the intervals:

1. $x < -5$: Use $x = -6$. $(-6 + 5)(-6 - 3) = 9 > 0$. This interval satisfies the inequality.
2. $-5 < x < 3$: Use $x = 0$. $(0 + 5)(0 - 3) = -15 < 0$. This interval does not satisfy the inequality.
3. $x > 3$: Use $x = 4$. $(4 + 5)(4 - 3) = 9 > 0$. This interval satisfies the inequality.

The solution for B is $x < -5$ or $x > 3$.