(4x^2-4x+1)(x^2+6x+5)《0

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вопрос

To solve this inequality, we first find the roots of both quadratic equations within the brackets:

For (4x^2-4x+1):

We can factor it as (2x-1)^2 which means there is a repeated root at x = 1/2.

For (x^2+6x+5):

We can factor it as (x+5)(x+1) which means the roots are x = -5 and x = -1.

Now, we can determine the sign of each quadratic in the intervals between the roots:

Let's consider the intervals x < -5, -5 < x < -1, -1 < x < 1/2, and x > 1/2.

Within these intervals, the signs of both quadratics are as follows:
(4x^2-4x+1) is always positive (since it is a square term)
(x^2+6x+5) is positive for x < -5 and x > -1, and negative for -5 < x < -1

To satisfy the inequality (4x^2-4x+1)(x^2+6x+5) < 0, we are interested in the intervals where one quadratic is negative and the other is positive.

The solution is:
-5 < x < -1

Therefore, the inequality (4x^2-4x+1)(x^2+6x+5) < 0 is satisfied when -5 < x < -1.