Question

question
solve the following inequality algebraically.
$x^2 + x - 30 \leq 0$
answer attempt 1 out of 2

Explanation:

Step1: Factor the quadratic

We factor \(x^{2}+x - 30\). We need two numbers that multiply to \(- 30\) and add to \(1\). The numbers are \(6\) and \(-5\), so \(x^{2}+x - 30=(x + 6)(x - 5)\). So the inequality becomes \((x + 6)(x - 5)\leq0\).

Step2: Find critical points

Set each factor equal to zero: \(x+6 = 0\) gives \(x=-6\); \(x - 5=0\) gives \(x = 5\). 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., \(x=-7\)): \((-7 + 6)(-7 - 5)=(-1)(-12)=12>0\), so the inequality is not satisfied.
• For \(-6
• For \(x>5\) (e.g., \(x = 6\)): \((6 + 6)(6 - 5)=(12)(1)=12>0\), so the inequality is not satisfied.

Step4: Check critical points

At \(x=-6\): \((-6 + 6)(-6 - 5)=(0)(-11)=0\leq0\), so \(x=-6\) is included. At \(x = 5\): \((5 + 6)(5 - 5)=(11)(0)=0\leq0\), so \(x = 5\) is included.

Answer:

\([-6,5]\)