Question

1.X²+4x-96>0 2.x²+3x-28<=0 3.x²-16x+64>=0

Answer 1

Ответ:

1.x1=8; x2=-12

2.найди D

D=9+4*28=121= 11 кв

х1=(-3+11)\2= 4

х2=(-3-11)\2= -7

2. приравниваем к нулю :

x^2+ 16x + 64 = 0  

a = 1 ; b=16 ; c=64

b^2 - 4ac = 16^2 - 4*1*64 = 256 - 256 = 0.

x1 = -16+0/2*1 = -8

x2 = -16-0/2*1 = -8

Answer 2

To solve these quadratic inequalities, let's analyze each one separately:

1. x² + 4x - 96 > 0: To solve this inequality, we can find the roots of the quadratic equation x² + 4x - 96 = 0. We can factorize it or use the quadratic formula to find the roots. Factoring, we have: (x + 12)(x - 8) > 0

Now, we can create a sign chart to determine the regions where the inequality is satisfied:

luaCopy code
  -12    8
   ---|---|---  
     - 0 +

We observe that the inequality is satisfied when x is in the interval (-∞, -12) or (8, +∞).

2. x² + 3x - 28 ≤ 0: Similarly, we find the roots of the quadratic equation x² + 3x - 28 = 0. Factoring, we have: (x + 7)(x - 4) ≤ 0

Creating a sign chart:

luaCopy code
  -7    4
   ---|---|---  
     + 0 -

Here, the inequality is satisfied when x is in the interval [-7, 4].

3. x² - 16x + 64 ≥ 0: Finding the roots of x² - 16x + 64 = 0, we can factorize it: (x - 8)(x - 8) ≥ 0

The equation simplifies to (x - 8)² ≥ 0, which is always true since the square of a real number is always non-negative. Therefore, this inequality is satisfied for all real values of x.

In summary:

1. x ∈ (-∞, -12) U (8, +∞)
2. x ∈ [-7, 4]
3. x ∈ (-∞, +∞)