Question

question
solve the following inequality algebraically.
$x^2 + 18x + 80 \leq 0$
answer attempt 1 out of 3
submit answer
∅ or ∪ ∞ < > ≤ ≥ , (,) {,} (,) all real numbers

Explanation:

Step1: Factor the quadratic expression

To solve \(x^{2}+18x + 80\leq0\), first factor the quadratic. We need two numbers that multiply to \(80\) and add to \(18\). Those numbers are \(8\) and \(10\). So, \(x^{2}+18x + 80=(x + 8)(x+10)\). The inequality becomes \((x + 8)(x + 10)\leq0\).

Step2: Find the critical points

Set each factor equal to zero: \(x+8 = 0\) gives \(x=-8\); \(x + 10=0\) gives \(x=-10\). These critical points divide the number line into three intervals: \((-\infty,-10)\), \((-10,-8)\), and \((-8,\infty)\).

Step3: Test intervals

• For \(x<-10\) (e.g., \(x = - 11\)): \((-11 + 8)(-11+10)=(-3)(-1)=3>0\), so the inequality is not satisfied here.
• For \(-10
• For \(x>-8\) (e.g., \(x = 0\)): \((0 + 8)(0+10)=(8)(10)=80>0\), so the inequality is not satisfied here.

Step4: Check critical points

Check \(x=-10\): \((-10 + 8)(-10+10)=(-2)(0)=0\leq0\), so \(x = - 10\) is included.
Check \(x=-8\): \((-8 + 8)(-8+10)=(0)(2)=0\leq0\), so \(x=-8\) is included.

Answer:

\(-10\leq x\leq - 8\) (or in interval notation \([-10,-8]\))