Question

practice for exam 4
2 of 22
this quiz: 22 point(s) possible
this question: 1 point(s) possible
solve the following inequality.
$x^2 + 3x + 2 < 0$
select the correct choice below and, if necessary, fill in the answer box.
a. the solution set is \boxed{}.
( type your answer in interval notation. use integers or fractions for any numbers in the expression. )
b. there is no real solution.

Explanation:

Step1: Factor the quadratic

Factor \(x^2 + 3x + 2\) as \((x + 1)(x + 2)\), so the inequality becomes \((x + 1)(x + 2) < 0\).

Step2: Find critical points

Set each factor to zero: \(x + 1 = 0\) gives \(x = -1\), \(x + 2 = 0\) gives \(x = -2\). These divide the number line into intervals: \((-\infty, -2)\), \((-2, -1)\), \((-1, \infty)\).

Step3: Test intervals

• For \(x < -2\) (e.g., \(x = -3\)): \((-3 + 1)(-3 + 2)=(-2)(-1)=2 > 0\), not less than 0.
• For \(-2 < x < -1\) (e.g., \(x = -1.5\)): \((-1.5 + 1)(-1.5 + 2)=(-0.5)(0.5)= -0.25 < 0\), satisfies the inequality.
• For \(x > -1\) (e.g., \(x = 0\)): \((0 + 1)(0 + 2)=(1)(2)=2 > 0\), not less than 0.

Step4: Write the solution

The solution is the interval where the inequality holds, which is \((-2, -1)\).

Answer:

A. The solution set is \((-2, -1)\)