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question 1 (multiple choice worth 5 points) (07.01r mc) what is the solution to $x^2 - 9x < -18$? $x < -6$ or $x > 3$ $-6 < x < 3$ $x < 3$ or $x > 6$ $3 < x < 6$
Step1: Rearrange the inequality
First, we rewrite the inequality \(x^{2}-9x < - 18\) as \(x^{2}-9x + 18<0\).
Step2: Factor the quadratic
Factor the quadratic expression \(x^{2}-9x + 18\). We need two numbers that multiply to \(18\) and add up to \(-9\). The numbers are \(-3\) and \(-6\). So, \(x^{2}-9x + 18=(x - 3)(x - 6)\). The inequality becomes \((x - 3)(x - 6)<0\).
Step3: Find critical points
The critical points are the values of \(x\) that make the expression equal to zero, which are \(x = 3\) and \(x=6\). These points divide the number line into three intervals: \((-\infty,3)\), \((3,6)\), and \((6,\infty)\).
Step4: Test intervals
- For the interval \((-\infty,3)\), let's choose \(x = 0\). Then \((0 - 3)(0 - 6)=(-3)\times(-6) = 18>0\), so the inequality is not satisfied.
- For the interval \((3,6)\), let's choose \(x = 4\). Then \((4 - 3)(4 - 6)=(1)\times(-2)=-2<0\), so the inequality is satisfied.
- For the interval \((6,\infty)\), let's choose \(x = 7\). Then \((7 - 3)(7 - 6)=(4)\times(1) = 4>0\), so the inequality is not satisfied.
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\(3 < x < 6\) (corresponding to the option: \(3 < x < 6\))