Question

(3х+1)^2+33>(2х+5)^2​

Answer 1

Ответ в файле

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Answer 2

Let's simplify and solve the inequality step by step:

Step 1: Expand the expressions within the parentheses. (3x + 1)^2 + 33 > (2x + 5)^2

Expanding the square terms: (3x + 1)(3x + 1) + 33 > (2x + 5)(2x + 5) 9x^2 + 6x + 1 + 33 > 4x^2 + 20x + 25

Step 2: Move all terms to one side of the inequality to get it in the standard form (ax^2 + bx + c > 0): 9x^2 + 6x + 1 + 33 - 4x^2 - 20x - 25 > 0

Step 3: Combine like terms: 5x^2 - 14x + 9 > 0

Now, we have a quadratic inequality in the form of ax^2 + bx + c > 0. To solve this inequality, we can use either the graphing method or the algebraic method (factoring and solving for x). Let's use the algebraic method:

Step 4: Factor the quadratic expression (if possible): The expression 5x^2 - 14x + 9 cannot be factored further using integer coefficients, so we move on to the next step.

Step 5: Solve for x: Since the inequality is in the form ax^2 + bx + c > 0, we want to find the intervals where the parabola is above the x-axis (positive) to satisfy the inequality.

To do this, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 5, b = -14, and c = 9:

x = (14 ± √((-14)^2 - 4 * 5 * 9)) / 2 * 5 x = (14 ± √(196 - 180)) / 10 x = (14 ± √16) / 10 x = (14 ± 4) / 10

Now we have two potential solutions for x:

1. x = (14 + 4) / 10 = 18 / 10 = 9 / 5
2. x = (14 - 4) / 10 = 10 / 10 = 1

Step 6: Determine the intervals where the inequality is satisfied: Now we need to test the inequality in the intervals to see where it holds true.

Test x = 0 (a value less than 1): 5(0)^2 - 14(0) + 9 = 9 > 0 (True)

Test x = 1 (a value between 1 and 9/5): 5(1)^2 - 14(1) + 9 = 0 (Not greater than 0)

Test x = 2 (a value between 9/5 and 3): 5(2)^2 - 14(2) + 9 = -3 (Not greater than 0)

Test x = 3 (a value greater than 3): 5(3)^2 - 14(3) + 9 = 9 > 0 (True)

Step 7: Final solution: The inequality is true when x is either less than 1 or greater than 3: x < 1 or x > 3.

Therefore, the solution to the inequality (3x + 1)^2 + 33 > (2x + 5)^2 is x < 1 or x > 3.