Question

3x^2 ≥ | x^2+2x | + 12

Answer 1

$3x^2\geq |x^2+2x|+12$

$\left \{ {{x^2+2x\geq0} \atop {3x^2\geq x^2+2x+12}} \right.$ 

$\left \{ {{x(x+2)\geq0} \atop {x^2-x-6\geq 0}} \right.$

$\left \{ {{x(-\infty;-2]\cup[0;+\infty)} \atop {x^2-x-6\geq 0}} \right.$

D = 25

x = 3

x = -2

  

$\left \{ {{x^2+2x$ 

$\left \{ {{x(x+2)$

$\left \{ {{x(-2;0)} \atop {2x^2-x-6\geq 0}} \right.$

D = 49

x = 3/2 - не принадлежит области определения

x = -2

 

Ответ: x = -2; x = 3 

 

Answer 2

To solve the inequality 3x^2 ≥ |x^2 + 2x| + 12, we need to consider two cases: when x^2 + 2x is positive and when it is negative.

Case 1: x^2 + 2x ≥ 0 In this case, the absolute value simplifies to x^2 + 2x. So the inequality becomes: 3x^2 ≥ x^2 + 2x + 12

Simplifying further, we have: 2x^2 - 2x - 12 ≥ 0

Next, we solve the quadratic equation by factoring or using the quadratic formula: 2x^2 - 2x - 12 = 0

Factoring out a 2, we get: 2(x^2 - x - 6) = 0

Now, factor the quadratic equation: 2(x - 3)(x + 2) = 0

Setting each factor equal to zero, we have: x - 3 = 0 => x = 3 x + 2 = 0 => x = -2

So in this case, the solutions are x = 3 and x = -2.

Case 2: x^2 + 2x < 0 In this case, the absolute value simplifies to -(x^2 + 2x). So the inequality becomes: 3x^2 ≥ -(x^2 + 2x) + 12

Simplifying further, we have: 3x^2 ≥ -x^2 - 2x + 12

Combining like terms, we get: 4x^2 + 2x - 12 ≥ 0

Now, we solve the quadratic equation by factoring or using the quadratic formula: 4x^2 + 2x - 12 = 0

Dividing the equation by 2, we have: 2x^2 + x - 6 = 0

Factoring the quadratic equation, we get: (2x - 3)(x + 2) = 0

Setting each factor equal to zero, we have: 2x - 3 = 0 => x = 3/2 x + 2 = 0 => x = -2

So in this case, the solutions are x = 3/2 and x = -2.

In conclusion, the solutions to the original inequality 3x^2 ≥ |x^2 + 2x| + 12 are x = 3, x = -2, x = 3/2.