Question

Квадратные неравенства :Решите : 1)3x^2-5x+6<02)-x^2+12x-36<03)2x^2-4x+13>0Заранее спасибо !)

Answer 1

1)3x²-5x+6<0
D=25-72=-47<0-нет решения,т.к. при любом х 3x²-5x+6Ю0
2)-x²+12x-36<0
x²-12x+36>0
(x-6)²>0
x∈(-∞;6) U (6;∞)
3)2x²-4x+13>0
D=16-104=-88<0
x∈(-∞;∞)

Answer 2

Quadratic Inequalities:

To solve quadratic inequalities, we need to find the values of x that satisfy the given inequality. Let's solve each of the quadratic inequalities you provided one by one:

1) 3x^2 - 5x + 6 < 0:

To solve this inequality, we can start by finding the roots of the quadratic equation 3x^2 - 5x + 6 = 0. The roots can be found using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the given equation, a = 3, b = -5, and c = 6. Plugging these values into the quadratic formula, we get:

x = (-(-5) ± √((-5)^2 - 4 * 3 * 6)) / (2 * 3) x = (5 ± √(25 - 72)) / 6 x = (5 ± √(-47)) / 6

Since the discriminant (√(b^2 - 4ac)) is negative, the quadratic equation has no real roots. Therefore, the quadratic expression 3x^2 - 5x + 6 is always positive or zero.

Hence, the solution to the inequality 3x^2 - 5x + 6 < 0 is no solution.

2) -x^2 + 12x - 36 < 0:

To solve this inequality, we can start by finding the roots of the quadratic equation -x^2 + 12x - 36 = 0. The roots can be found using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the given equation, a = -1, b = 12, and c = -36. Plugging these values into the quadratic formula, we get:

x = (-(12) ± √((12)^2 - 4 * (-1) * (-36))) / (2 * (-1)) x = (-12 ± √(144 - 144)) / (-2) x = (-12 ± √0) / (-2) x = (-12 ± 0) / (-2) x = -12 / (-2) x = 6

Since the discriminant (√(b^2 - 4ac)) is zero, the quadratic equation has one real root. Therefore, the quadratic expression -x^2 + 12x - 36 is always negative or zero when x = 6.

Hence, the solution to the inequality -x^2 + 12x - 36 < 0 is x < 6.

3) 2x^2 - 4x + 13 > 0:

To solve this inequality, we can start by finding the roots of the quadratic equation 2x^2 - 4x + 13 = 0. The roots can be found using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the given equation, a = 2, b = -4, and c = 13. Plugging these values into the quadratic formula, we get:

x = (-(-4) ± √((-4)^2 - 4 * 2 * 13)) / (2 * 2) x = (4 ± √(16 - 104)) / 4 x = (4 ± √(-88)) / 4

Since the discriminant (√(b^2 - 4ac)) is negative, the quadratic equation has no real roots. Therefore, the quadratic expression 2x^2 - 4x + 13 is always positive or zero.

Hence, the solution to the inequality 2x^2 - 4x + 13 > 0 is all real numbers.

I hope this helps! Let me know if you have any further questions.