Question

1)2x^2+3x-9>02)5x-x^2<03)(x+3)(1+x)>04)\frac{x-3}{x-1}<05)eft \{ {{3-4x<14} \atop {1-2x>0}} \right.ПОМОГИТЕ!!!!!!

Answer 1

2x^2+3x-9>0

a=2,b=3,c=9 

D=b^2-4ac=9-4*(-9)*2=9+72=81

x1=-b-корень из D/2a=-3-9/4=-3

x2=-b+корень из D/2a=-3+9/4=1целая 1/2

Конец первого!

 

Answer 2

1) 2x^2 + 3x - 9 > 0

To solve the inequality 2x^2 + 3x - 9 > 0, we can use the quadratic formula or factorization method. Let's use the factorization method:

1. Factor the quadratic expression: 2x^2 + 3x - 9 = (2x - 3)(x + 3). 2. Set each factor equal to zero and solve for x: - 2x - 3 = 0 --> 2x = 3 --> x = 3/2 - x + 3 = 0 --> x = -3

Now we have the critical points x = 3/2 and x = -3. We can use these points to divide the number line into three intervals: (-∞, -3), (-3, 3/2), and (3/2, +∞).

To determine the sign of the expression 2x^2 + 3x - 9 in each interval, we can choose a test point from each interval and substitute it into the expression. Let's choose -4, 0, and 2 as test points:

- For x = -4: 2(-4)^2 + 3(-4) - 9 = 32 - 12 - 9 = 11. Since the result is positive, the expression is positive in the interval (-∞, -3). - For x = 0: 2(0)^2 + 3(0) - 9 = -9. Since the result is negative, the expression is negative in the interval (-3, 3/2). - For x = 2: 2(2)^2 + 3(2) - 9 = 8 + 6 - 9 = 5. Since the result is positive, the expression is positive in the interval (3/2, +∞).

Therefore, the solution to the inequality 2x^2 + 3x - 9 > 0 is x ∈ (-∞, -3) ∪ (3/2, +∞). [[1]]

2) 5x - x^2 < 0

To solve the inequality 5x - x^2 < 0, we can factorize the quadratic expression:

1. Rewrite the inequality: -x^2 + 5x < 0. 2. Factor out -x: -x(x - 5) < 0. 3. Now we have two factors: -x and (x - 5). 4. Set each factor equal to zero and solve for x: - -x = 0 --> x = 0 - x - 5 = 0 --> x = 5

Now we have the critical points x = 0 and x = 5. We can use these points to divide the number line into three intervals: (-∞, 0), (0, 5), and (5, +∞).

To determine the sign of the expression -x(x - 5) in each interval, we can choose a test point from each interval and substitute it into the expression. Let's choose -1, 1, and 6 as test points:

- For x = -1: -(-1)(-1 - 5) = 6. Since the result is positive, the expression is positive in the interval (-∞, 0). - For x = 1: -(1)(1 - 5) = 4. Since the result is negative, the expression is negative in the interval (0, 5). - For x = 6: -(6)(6 - 5) = -6. Since the result is negative, the expression is negative in the interval (5, +∞).

Therefore, the solution to the inequality 5x - x^2 < 0 is x ∈ (-∞, 0) ∪ (5, +∞). [[2]]

3) (x + 3)(1 + x) > 0

To solve the inequality (x + 3)(1 + x) > 0, we can use the zero-product property:

1. Set each factor equal to zero and solve for x: - x + 3 = 0 --> x = -3 - 1 + x = 0 --> x = -1

Now we have the critical points x = -3 and x = -1. We can use these points to divide the number line into three intervals: (-∞, -3), (-3, -1), and (-1, +∞).

To determine the sign of the expression (x + 3)(1 + x) in each interval, we can choose a test point from each interval and substitute it into the expression. Let's choose -4, -2, and 0 as test points:

- For x = -4: (-4 + 3)(1 + (-4)) = (-1)(-3) = 3. Since the result is positive, the expression is positive in the interval (-∞, -3). - For x = -2: (-2 + 3)(1 + (-2)) = (1)(-1) = -1. Since the result is negative, the expression is negative in the interval (-3, -1). - For x = 0: (0 + 3)(1 + 0) = (3)(1) = 3. Since the result is positive, the expression is positive in the interval (-1, +∞).

Therefore, the solution to the inequality (x + 3)(1 + x) > 0 is x ∈ (-∞, -3) ∪ (-1, +∞). [[3]]

4) (x - 3)/(x - 1) < 0

To solve the inequality (x - 3)/(x - 1) < 0, we can use the sign chart method:

1. Set the numerator and denominator equal to zero and solve for x: - x - 3 = 0 --> x = 3 - x - 1 = 0 --> x = 1

Now we have the critical points x = 3 and x = 1. We can use these points to divide the number line into three intervals: (-∞, 1), (1, 3), and (3, +∞).

To determine the sign of the expression (x - 3)/(x - 1) in each interval, we can choose a test point from each interval and substitute it into the expression. Let's choose 0, 2, and 4 as test points:

- For x = 0: (0 - 3)/(0 - 1) = 3/(-1) = -3. Since the result is negative, the expression is negative in the interval (-∞, 1). - For x = 2: (2 - 3)/(2 - 1) = -1/1 = -1. Since the result is negative, the expression is negative in the interval (1, 3). - For x = 4: (4 - 3)/(4 - 1) = 1/3. Since the result is positive, the expression is positive in the interval (3, +∞).

Therefore, the solution to the inequality (x - 3)/(x - 1) < 0 is x ∈ (-∞, 1) ∪ (3, +∞). [[4]]

5) left { {3 - 4x < 14} over {1 - 2x > 0} right }

The given expression is a system of inequalities. Let's solve each inequality separately:

- 3 - 4x < 14: 1. Subtract 3 from both sides: -4x < 11. 2. Divide both sides by -4, remembering to reverse the inequality sign since we are dividing by a negative number: x > -11/4.

- 1 - 2x > 0: 1. Subtract 1 from both sides: -2x > -1. 2. Divide both sides by -2, remembering to reverse the inequality sign: x < 1/2.

Therefore, the solution to the system of inequalities is x ∈ (-11/4, 1/2).