Question

(х^2-3x-2)(х^2-3x+1)<10

Answer 1

(x² -3x-2)(x²-3x+1) < 10
Пусть y=x² -3x
(y-2)(y+1)-10 < 0
y² -2y+y-2-10 <0
y² - y -12 < 0
f(y) = y² -y-12 - парабола, ветви вверх.
y² -y-12=0
D=1-4*(-12)=1+48=49
y₁ = 1-7 = -3
         2
y₂ = 1+7 = 4
         2
     +                    -                 +
-------- -3 ----------- 4 ------------
                \\\\\\\\\\\\\\
y∈(-3; 4)
-3 < y < 4

{x² -3x> -3
{x² -3x<4

x² -3x>-3
x² -3x+3>0
f(x)=x²-3x+3 - парабола, ветви вверх.
D= 9-4*3= -3 < 0
Парабола не пересекает ось ОХ.
Парабола лежит выше оси ОХ.
Неравенство выполняется при любом Х.
х∈(-∞; +∞)

x² -3x <4
x² -3x-4<0
f(x)=x² -3x-4 - парабола, ветви вверх.
x² -3x-4=0
D=9-4*(-4)=9+16=25
x₁ = 3-5 =  -1
          2
x₂ = 3+5 =4
         2
      +                  -                 +
------- -1 ----------- 4 -------------
                \\\\\\\\\\\\
x∈(-1; 4)

Ответ: (-1; 4)

Answer 2

To solve the inequality $(x^2 - 3x - 2)(x^2 - 3x + 1) < 10$, you can follow these steps:

1. Expand the left side of the inequality: $(x^2 - 3x - 2)(x^2 - 3x + 1) = x^4 - 3x^3 - 2x^2 - 3x^3 + 9x^2 + 6x - 2x^2 + 6x - 2$

2. Combine like terms: $x^4 - 6x^3 + 7x^2 + 12x - 2 < 10$

3. Move all terms to the left side of the inequality: $x^4 - 6x^3 + 7x^2 + 12x - 12 < 0$

4. Factor the expression: $(x - 2)(x + 1)(x^2 - 3x + 6) < 0$

Now, you need to find the intervals where this expression is less than zero (negative). To do this, you can use a sign chart or test values in each interval.

1. The factor $x - 2$ changes sign at $x = 2$.
2. The factor $x + 1$ changes sign at $x = -1$.
3. The quadratic factor $x^2 - 3x + 6$ is always positive because its discriminant ($b^2 - 4ac$) is negative, which means it doesn't have real roots.

Now, create a sign chart for these factors:

Interval	x - 2	x + 1	x^2 - 3x + 6	Expression
x < -1	-	-	+	-
-1 < x < 2	-	+	+	-
x > 2	+	+	+	+

The expression is negative in the intervals $-1 < x < 2$. Therefore, the solution to the inequality $(x^2 - 3x - 2)(x^2 - 3x + 1) < 10$ is:

$-1 < x < 2$