The solution set of 2x^2+6x+4<0

Answer 1

To solve the inequality 2x^2+6x+4<0, we can follow these steps:

1. Divide both sides by 2 to simplify the coefficients:
x^2 + 3x + 2 < 0

2. Factor the quadratic expression:
(x + 1)(x + 2) < 0

3. Find the critical values of x where the expression changes signs:
x + 1 = 0 ⇒ x = -1
x + 2 = 0 ⇒ x = -2

4. Create a sign chart and test the expression in each interval:
- + -
|____|____|____|
-2 -1 0

For x < -2, both factors are negative, so their product is positive.
For -2 < x < -1, only (x+2) is positive, so the product is negative.
For x > -1, both factors are positive, so their product is positive.

5. Write the solution set by combining the intervals where the expression is negative:
x ∈ (-2, -1)

Therefore, the solution set of 2x^2+6x+4<0 is the open interval (-2, -1).

Answer 2

2x^2+6x+4<0
x^2 + 3x + 2 < 0
(x+2)(x+1) < 0

-2 < x < -1

Answer 3

You are correct! The solution set of the inequality (x+2)(x+1) < 0 is (-2, -1), which means that -2 < x < -1 is the set of values of x that make the inequality true.