Question

solve for x.
$4x^{2} + 4x - 24 > 0$
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3.

Explanation:

Step1: Simplify the inequality

First, divide the entire inequality \(4x^{2}+4x - 24>0\) by 4 to simplify it. We get \(x^{2}+x - 6>0\).

Step2: Factor the quadratic

Factor the quadratic expression \(x^{2}+x - 6\). We need two numbers that multiply to -6 and add to 1. Those numbers are 3 and -2. So, \(x^{2}+x - 6=(x + 3)(x - 2)\). The inequality becomes \((x + 3)(x - 2)>0\).

Step3: Find critical points

Set each factor equal to zero to find the critical points: \(x+3 = 0\) gives \(x=-3\) and \(x - 2=0\) gives \(x = 2\). These critical points divide the number line into three intervals: \((-\infty,-3)\), \((-3,2)\), and \((2,\infty)\).

Step4: Test intervals

• For the interval \((-\infty,-3)\), let's choose \(x=-4\). Substitute into \((x + 3)(x - 2)\): \((-4 + 3)(-4 - 2)=(-1)\times(-6)=6>0\), so this interval is part of the solution.
• For the interval \((-3,2)\), let's choose \(x = 0\). Substitute into \((x + 3)(x - 2)\): \((0 + 3)(0 - 2)=(3)\times(-2)=-6<0\), so this interval is not part of the solution.
• For the interval \((2,\infty)\), let's choose \(x = 3\). Substitute into \((x + 3)(x - 2)\): \((3 + 3)(3 - 2)=(6)\times(1)=6>0\), so this interval is part of the solution.

Answer:

\(x < - 3\) or \(x>2\)