Question

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

Explanation:

Step1: Factor the quadratic

We factor \(x^2 - x - 30\). 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 \(x = 6\), and \(x + 5 = 0\) gives \(x=-5\). These critical points divide the number line into three intervals: \((-\infty, -5)\), \((-5, 6)\), and \((6, \infty)\).

Step3: Test intervals

• For \(x < -5\) (e.g., \(x=-6\)): \((-6 - 6)(-6 + 5)=(-12)(-1)=12>0\), so the inequality is not satisfied.
• For \(-5 < x < 6\) (e.g., \(x = 0\)): \((0 - 6)(0 + 5)=(-6)(5)=-30\leq0\), so the inequality is satisfied.
• For \(x > 6\) (e.g., \(x = 7\)): \((7 - 6)(7 + 5)=(1)(12)=12>0\), so the inequality is not satisfied.

Step4: Include critical points

Since the inequality is \(\leq0\), we include the critical points \(x=-5\) and \(x = 6\) because when \(x=-5\) or \(x = 6\), \((x - 6)(x + 5)=0\) which satisfies the inequality.

Answer:

\(-5\leq x\leq6\)