Question

solve the following quadratic inequality and wr
the quadratic is factorable so you should try th
graphing.
$x^2 - 8x + 12 < 0$
note: use oo for $\infty$ and u for union
question help: message instructor
check answer

Explanation:

Step1: Factor the quadratic

We factor \(x^{2}-8x + 12\). We need two numbers that multiply to \(12\) and add up to \(- 8\). The numbers are \(-6\) and \(-2\). So, \(x^{2}-8x + 12=(x - 6)(x - 2)\)

Step2: Find the critical points

Set each factor equal to zero: \(x-6 = 0\) gives \(x = 6\) and \(x - 2=0\) gives \(x=2\)

Step3: Test intervals

The critical points \(x = 2\) and \(x=6\) divide the number line into three intervals: \((-\infty,2)\), \((2,6)\) and \((6,\infty)\)

• For the interval \((-\infty,2)\), let's test \(x = 0\): \((0 - 6)(0 - 2)=(-6)\times(-2) = 12>0\), so the inequality \((x - 6)(x - 2)<0\) is not satisfied.
• For the interval \((2,6)\), let's test \(x = 3\): \((3 - 6)(3 - 2)=(-3)\times(1)=- 3<0\), so the inequality is satisfied.
• For the interval \((6,\infty)\), let's test \(x = 7\): \((7 - 6)(7 - 2)=(1)\times(5) = 5>0\), so the inequality is not satisfied.

Answer:

\(2 < x < 6\) or in the required format \((2,6)\) (since the inequality is strict, we use open intervals)