Question

-2хв квадрате-3х+2<0​

Answer 1

Ответ:

Пошаговое объяснение:

-2x^2-3x+2<0

2x^2+3x-2>0

(x+2)(2x-1)>0

(x+2)(x-1/2)>0

x>1/2

x<-2

x€(-inf;-2)U(1/2;inf)

€ - принадлежит

inf - бесконечность

Answer 2

To solve the quadratic inequality $-2x^2 - 3x + 2 < 0$, you can follow these steps:

Step 1: Find the critical points by setting the quadratic expression equal to zero and solving for $x$:

$-2x^2 - 3x + 2 = 0$

You can solve this equation using the quadratic formula:

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

For the given quadratic $-2x^2 - 3x + 2$, $a = -2$, $b = -3$, and $c = 2$. Plugging these values into the formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-2)(2)}}{2(-2)}$ $x = \frac{3 \pm \sqrt{9 + 16}}{-4}$ $x = \frac{3 \pm \sqrt{25}}{-4}$ $x = \frac{3 \pm 5}{-4}$

This gives two possible values for $x$:

$x_1 = -2$ $x_2 = \frac{1}{2}$

Step 2: Plot these critical points on a number line:

Copy code
  -2             1/2
   ─────|───────|───────

Step 3: Test the intervals created by these critical points using a test point in each interval to determine where the quadratic expression is positive or negative.

Choose a value less than -2, say $x = -3$, and plug it into the quadratic expression:

$-2(-3)^2 - 3(-3) + 2 = -18 + 9 + 2 = -7$

Since this result is negative, the quadratic expression is negative in the interval $(-\infty, -2)$.

Choose a value between -2 and $\frac{1}{2}$, say $x = 0$, and plug it into the quadratic expression:

$-2(0)^2 - 3(0) + 2 = 2$

Since this result is positive, the quadratic expression is positive in the interval $(-2, 1/2)$.

Step 4: Determine the solution based on the intervals where the expression is less than zero:

The solution to the inequality $-2x^2 - 3x + 2 < 0$ is the interval $(-\infty, -2)$.

Therefore, the solution to the inequality is:

$x \in (-\infty, -2)$