Question

2x²-3x+4>0Ппжжжжпжпжажпжжааажжппжаж​

Answer 1

Ответ:

х∈(-∞;+∞)

Пошаговое объяснение:

2x²-3x+4>0

D=9-4*2*4<0

решаем другим способом

находим вершину параболы. т.к. а>0, то ветви направлены вверх

х=-b/2a

x=3/4

y=2*(3/4)²-3*(3/4)+4=2 7/8

y>0, значит

неравенство имеет бесконечное множество решений. При любом значении х неравенство будет больше 0.

х∈(-∞;+∞)

Answer 2

It seems like you've provided an inequality in an unusual format, including characters like "Ппжжжжпжпжажпжжааажжппжаж" which don't belong to standard mathematical notation. However, I'll interpret the quadratic inequality for you and solve it.

The given inequality is $2x^2 - 3x + 4 > 0$.

To solve this inequality, we'll first find the critical points where the expression equals zero ($2x^2 - 3x + 4 = 0$), and then determine the regions where the expression is greater than zero.

Let's start by finding the critical points using the quadratic formula:

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $ax^2 + bx + c$ is the quadratic equation.

In this case, $a = 2$, $b = -3$, and $c = 4$. Plugging these values into the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 2 \times 4}}{2 \times 2}$

$x = \frac{3 \pm \sqrt{9 - 32}}{4}$

Since the discriminant $9 - 32$ is negative, the quadratic equation $2x^2 - 3x + 4 = 0$ does not have real roots.

Now, we'll determine the regions where $2x^2 - 3x + 4 > 0$.

Since the coefficient of $x^2$ is positive ($a = 2 > 0$), the parabola opens upwards. This means that the expression is greater than zero either when $x$ is less than both roots (which don't exist in this case) or when $x$ is greater than both roots.

Since there are no real roots, the expression $2x^2 - 3x + 4$ is always positive. Therefore, the solution to the inequality $2x^2 - 3x + 4 > 0$ is all real numbers $x$.

In mathematical notation, this can be represented as: $x \in \mathbb{R}$ (all real numbers).