Question

-x^2+3x-2<0 Решите неравенство ​

Answer 1

-x²+3x-2<0

x²-3x+2>0

x²-x-2x+2>0

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

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

x∈(-∞;1)∪(2;+∞)

Answer 2

To solve the inequality $-x^2 + 3x - 2 < 0$, we need to find the values of $x$ that satisfy the inequality. Here's how to do it:

Step 1: Find the critical points by setting the expression equal to zero and solving for $x$: $-x^2 + 3x - 2 = 0$

To solve this quadratic equation, you can use factoring or the quadratic formula. In this case, it's easier to factor it: $-(x - 2)(x - 1) = 0$

So, the critical points are $x = 1$ and $x = 2$.

Step 2: Create a sign chart to identify the intervals where the expression $-x^2 + 3x - 2$ is positive or negative.

Interval	-x^2 + 3x - 2
$x < 1$	$- \times - \times -$ (negative)
$1 < x < 2$	$+ \times - \times -$ (positive)
$x > 2$	$+ \times + \times -$ (negative)

Step 3: Identify the solution:

The solution to the inequality is the interval where the expression $-x^2 + 3x - 2$ is less than zero (negative). From the sign chart, we can see that this happens when $1 < x < 2$.

Therefore, the solution to the inequality $-x^2 + 3x - 2 < 0$ is $1 < x < 2$.