Question

1. 5 -x2 + 3x – 4 < 0. 3. 5 x2 + 12x > -36.2. 5 (2x + 3)(x - 1)(5 + x)<0.4. 6 (x2 – 5x + 6)(x2 – 9)<0.​

Answer 1

Ответ:

Вроде бы так только вот я не помню со школы - х что то переноситься или нет так чтт вроде бы так

Answer 2

To solve these inequalities, we need to find the values of x that satisfy each inequality. Let's solve them one by one:

1. 5 - x^2 + 3x - 4 < 0: Simplifying the inequality, we get: -x^2 + 3x + 1 < 0

   To solve this inequality, we can find the critical points by setting the expression equal to zero: -x^2 + 3x + 1 = 0

   We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

   In this case, a = -1, b = 3, and c = 1: x = (-(3) ± √((3)^2 - 4(-1)(1))) / (2(-1)) x = (-3 ± √(9 + 4)) / (-2) x = (-3 ± √13) / (-2)

   The critical points are x = (-3 + √13) / (-2) and x = (-3 - √13) / (-2).

   Now we need to determine the intervals where the expression is less than zero. We can use test points within each interval to determine the sign of the expression. Let's test the intervals (-∞, (-3 - √13) / (-2)), ((-3 - √13) / (-2), (-3 + √13) / (-2)), and ((-3 + √13) / (-2), +∞).

   For the interval (-∞, (-3 - √13) / (-2)), we can choose x = -4: -(-4)^2 + 3(-4) + 1 = -16 - 12 + 1 = -27 < 0

   For the interval ((-3 - √13) / (-2), (-3 + √13) / (-2)), we can choose x = 0: -(0)^2 + 3(0) + 1 = 1 > 0

   For the interval ((-3 + √13) / (-2), +∞), we can choose x = 2: -(2)^2 + 3(2) + 1 = 3 > 0

   Therefore, the solution to the inequality -x^2 + 3x + 1 < 0 is: (-∞, (-3 - √13) / (-2)) U ((-3 + √13) / (-2), +∞)

2. 5x^2 + 12x > -36: Rearranging the inequality, we get: 5x^2 + 12x + 36 > 0

   To solve this inequality, we can find the critical points by setting the expression equal to zero: 5x^2 + 12x + 36 = 0

   We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

   In this case, a = 5, b = 12, and c = 36: x = (-(12) ± √((12)^2 - 4(5)(36))) / (2(5)) x = (-12 ±