Question

1) X^2+4x+10>=0 2)x^2+10x+25>0 3)x^2+3x+=<0

Answer 1

X^2+4x+10>=0

Ответ:

x∈R или x - любое число.

Объяснение:

Решение квадратного уравнения

x^2+4x+10=0

Вычислим дискриминант.

D=b^2−4ac=−24

Ответ: корней нет, т.к. D<0

Т.к. a=1>0, то x^2+4x+10>0 для любых x

Т.к. корней уравнения x^2+4x+10=0 нет, то x^2+4x+10>0 для любых x

x^2+10x+25>0

Ответ:

x∈(−∞;−5)∪(−5;+∞) или x<−5; x>−5

Объяснение:

Решение квадратного уравнения

Вычислим дискриминант.

D=b^2−4ac=0

x1,2=(−b±√D)/2a=(−10±√0)/2=−5

Ответ: x1,2=−5

Корни уравнения

x:2+10x+25=0:

x1=−5

Наносим найденные точки на числовую ось и вычисляем знаки на каждом интервале (картинка в закрепе)

Answer 2

To solve these quadratic inequalities, we need to find the values of x that satisfy each inequality. We can start by finding the discriminant (the value inside the square root in the quadratic formula) for each equation to determine the nature of the roots.

1. $x^2 + 4x + 10 \geq 0$:

The discriminant, $D$, is given by $D = b^2 - 4ac$, where $a = 1$, $b = 4$, and $c = 10$.

$D = 4^2 - 4(1)(10) = 16 - 40 = -24$

Since the discriminant is negative ($D < 0$), the quadratic has no real roots, and it is always positive. Therefore, the solution to this inequality is all real numbers:

$x \in (-\infty, +\infty)$

2. $x^2 + 10x + 25 > 0$:

The discriminant, $D$, is given by $D = b^2 - 4ac$, where $a = 1$, $b = 10$, and $c = 25$.

$D = 10^2 - 4(1)(25) = 100 - 100 = 0$

Since the discriminant is zero ($D = 0$), the quadratic has one real root at $x = -\frac{b}{2a} = -\frac{10}{2} = -5$, and it is always non-negative.

So, the solution to this inequality is all real numbers except $x = -5$:

$x \in (-\infty, -5) \cup (-5, +\infty)$

3. $x^2 + 3x \leq 0$:

This inequality is already in the form of $ax^2 + bx + c \leq 0$, so we can find the solutions directly.

Factor the expression: $x(x + 3) \leq 0$

Now, we can analyze the sign of each factor:

• $x$ changes sign at $x = 0$.
• $x + 3$ changes sign at $x = -3$.

To satisfy $x(x + 3) \leq 0$, we need either $x$ and $x + 3$ to have opposite signs or one of them to be zero.

So, the solution is:

$x \in (-\infty, -3] \cup [0, +\infty)$

These are the solutions to the given quadratic inequalities.