Question

solve for x.
$x^2 + 11x + 30 \leq 0$
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3.

Explanation:

Step1: Factor the quadratic

We factor \(x^{2}+11x + 30\). We need two numbers that multiply to \(30\) and add to \(11\). Those numbers are \(5\) and \(6\). So, \(x^{2}+11x + 30=(x + 5)(x+6)\). The inequality becomes \((x + 5)(x + 6)\leq0\).

Step2: Find critical points

Set each factor equal to zero: \(x+5 = 0\) gives \(x=-5\); \(x + 6=0\) gives \(x=-6\). These critical points divide the number line into three intervals: \((-\infty,-6)\), \((-6,-5)\), and \((-5,\infty)\).

Step3: Test intervals

• For \(x<-6\) (e.g., \(x = -7\)): \((-7 + 5)(-7+6)=(-2)(-1)=2>0\), so the inequality is not satisfied.
• For \(-6
• For \(x>-5\) (e.g., \(x = 0\)): \((0 + 5)(0+6)=30>0\), so the inequality is not satisfied.
• At the critical points: When \(x=-6\), \((-6 + 5)(-6+6)=0\leq0\); when \(x=-5\), \((-5 + 5)(-5+6)=0\leq0\).

So the solution is \(-6\leq x\leq -5\).

Answer:

\(-6 \leq x \leq -5\)