
(2x+5)^2>=(5x-2)^2 решите пожалуйста в ответе должно получится -3/7 и 7/3 а у меня не получается


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Let's solve the inequality step by step to find the correct solution. We'll start by simplifying both sides of the inequality:
(2x + 5)^2 ≥ (5x - 2)^2
Expanding both sides:
(2x + 5)(2x + 5) ≥ (5x - 2)(5x - 2)
Now, apply the FOIL method (First, Outer, Inner, Last) to expand both sides:
4x^2 + 20x + 25 ≥ 25x^2 - 20x + 4
Now, bring all terms to one side of the inequality:
4x^2 + 20x + 25 - 25x^2 + 20x - 4 ≥ 0
Simplify:
-21x^2 + 40x + 21 ≥ 0
To solve this quadratic inequality, we need to find the critical points where the inequality may change direction. The critical points are the solutions to the equation -21x^2 + 40x + 21 = 0.
To find the critical points, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.
In our case, a = -21, b = 40, and c = 21:
x = [ -40 ± √(40^2 - 4(-21)(21)) ] / 2(-21) x = [ -40 ± √(1600 + 1764) ] / (-42) x = [ -40 ± √3364 ] / (-42) x = [ -40 ± 58 ] / (-42)
Now, we get two potential critical points:
- x = (-40 + 58) / (-42) = 18 / (-42) = -3 / 7
- x = (-40 - 58) / (-42) = -98 / (-42) = 7 / 3
Now, let's test the intervals created by these critical points. We can use a test point from each interval to determine the sign of -21x^2 + 40x + 21.
Test x = 0 (pick a value less than -3/7, e.g., x = -1):
-21(-1)^2 + 40(-1) + 21 = -21 + 40 + 21 = 40 > 0 (positive)
Test x = 1 (pick a value between -3/7 and 7/3):
-21(1)^2 + 40(1) + 21 = -21 + 40 + 21 = 40 > 0 (positive)
Test x = 2 (pick a value greater than 7/3, e.g., x = 3):
-21(3)^2 + 40(3) + 21 = -189 + 120 + 21 = -48 < 0 (negative)
The signs of -21x^2 + 40x + 21 in the intervals are:
- Negative for x < -3/7
- Positive for -3/7 < x < 7/3
- Negative for x > 7/3
We want to find where the inequality -21x^2 + 40x + 21 ≥ 0 holds. Thus, we are interested in the intervals where the expression is non-negative.
The solution is: -3/7 ≤ x ≤ 7/3.
So, your initial claim is incorrect. The correct solution to the inequality is -3/7 ≤ x ≤ 7/3.


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