Question

solve for x.
$4x^2 + 4x - 8 \geq 0$
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3.

Explanation:

Step1: Simplify the quadratic inequality

First, divide the entire inequality \(4x^{2}+4x - 8\geq0\) by 4 to simplify it. We get \(x^{2}+x - 2\geq0\).

Step2: Factor the quadratic expression

Factor the quadratic \(x^{2}+x - 2\). We need two numbers that multiply to -2 and add to 1. Those numbers are 2 and -1. So, \(x^{2}+x - 2=(x + 2)(x - 1)\). So the inequality becomes \((x + 2)(x - 1)\geq0\).

Step3: Find the critical points

Set each factor equal to zero: \(x+2 = 0\) gives \(x=-2\), and \(x - 1=0\) gives \(x = 1\). These critical points divide the number line into three intervals: \((-\infty,-2)\), \((-2,1)\), and \((1,\infty)\).

Step4: Test intervals

• For the interval \((-\infty,-2)\), let's pick \(x=-3\). Substitute into \((x + 2)(x - 1)\): \((-3 + 2)(-3 - 1)=(-1)\times(-4)=4\geq0\), so this interval is part of the solution.
• For the interval \((-2,1)\), let's pick \(x = 0\). Substitute into \((x + 2)(x - 1)\): \((0 + 2)(0 - 1)=(2)\times(-1)=-2<0\), so this interval is not part of the solution.
• For the interval \((1,\infty)\), let's pick \(x = 2\). Substitute into \((x + 2)(x - 1)\): \((2 + 2)(2 - 1)=(4)\times(1)=4\geq0\), so this interval is part of the solution.
• Also, we include the critical points \(x=-2\) and \(x = 1\) because the inequality is \(\geq0\).

Answer:

\(x\leq - 2\) or \(x\geq1\)