# Is there any value of x that satisfies x^2 + 3x + 2 >= 0 and 4x^2 + 5x + 1 >= 0

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### 1 Answer

The values of x that satisfy `x^2 + 3x + 2 >= 0` and `4x^2 + 5x + 1 >= 0` have to be determined.

`x^2 + 3x + 2 >= 0`

=> `x^2 + 2x + x + 2 >= 0`

=> `x(x + 2) +1(x + 2) >= 0`

=> `(x + 1)(x + 2) >= 0`

This is the case when:

- `x + 1 >= 0` and `x + 2 >= 0`

=> `x >= -1` and `x >= -2`

=> `x >= -1`

- `x + 1 <= 0` and `x + 2 <= 0`

=> `x <= -1` and `x <= -2`

=> `x <= -2`

The solution for this inequality is `(-oo, -2]U[-1, oo)`

`4x^2 + 5x + 1 >= 0`

=> `4x^2 + 4x + x + 1 >= 0`

=> `4x(x + 1) + 1(x + 1) >= 0`

=> `(4x + 1)(x + 1) >= 0`

This is true when

- `4x + 1 >= 0` and `x + 1 >= 0`

=> `x >= -1/4` and `x >= -1`

=> `x >= -1/4`

- `4x + 1 <= 0` and `x + 1 <= 0`

=> `x <= -1/4` and `x <= -1`

=> `x <= -1`

The solution of this inequality is `(-oo, -1]U[-1/4, oo)`

The intersection of `(-oo, -2]U[-1, oo)` and `(-oo, -1]U[-1/4, oo)` is `(-oo, -2]U[-1/4, oo)`

**The values of x that satisfy both the inequalities lie in **`(-oo, -2]U[-1/4, oo)`