Question

X^2+2^log2(-2x)-15<0

Answer 1

Решение прикреплено во вложениях ниже.

Answer 2

The given inequality is $x^2 + 2 \cdot 2^{\log_2(-2x)} - 15 < 0$.

Let's break down the expression step by step:

1. $2^{\log_2(-2x)}$ simplifies to $-2x$ because the logarithm base ($2$) and the exponent ($\log_2(-2x)$) cancel each other out.

So, the inequality becomes:

$x^2 + 2 \cdot (-2x) - 15 < 0$

Simplify further:

$x^2 - 4x - 15 < 0$

Now, we have a quadratic inequality. To solve it, we can factor the quadratic expression:

$(x - 5)(x + 3) < 0$

Now we need to find the intervals where the expression is less than zero. To do this, we'll look at the critical points (where the expression equals zero) and test points within the intervals determined by those critical points:

1. Critical points: $x = -3$ and $x = 5$.

2. Test point in the interval $(-∞, -3)$: Let's choose $x = -4$. $((-4) - 5) \cdot ((-4) + 3) = (-9) \cdot (-1) = 9 > 0$

   The expression is greater than zero in this interval.

3. Test point in the interval $(-3, 5)$: Let's choose $x = 0$. $(0 - 5) \cdot (0 + 3) = (-5) \cdot 3 = -15 < 0$

   The expression is less than zero in this interval.

4. Test point in the interval $(5, ∞)$: Let's choose $x = 6$. $(6 - 5) \cdot (6 + 3) = 1 \cdot 9 = 9 > 0$

   The expression is greater than zero in this interval.

To summarize, the expression $(x - 5)(x + 3)$ is less than zero in the interval $(-3, 5)$.

So, the solution to the inequality $x^2 + 2 \cdot 2^{\log_2(-2x)} - 15 < 0$ is $x$ belongs to the open interval $(-3, 5)$.