Question

solve for x.
$5x^2 + 20x - 60 \leq 0$
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 \(5x^{2}+20x - 60\leq0\) by 5 to simplify it. We get \(x^{2}+4x - 12\leq0\).

Step2: Factor the quadratic

Factor the quadratic expression \(x^{2}+4x - 12\). We need two numbers that multiply to - 12 and add up to 4. The numbers are 6 and - 2. So, \(x^{2}+4x - 12=(x + 6)(x - 2)\). The inequality becomes \((x + 6)(x - 2)\leq0\).

Step3: Find critical points

Set each factor equal to zero: \(x+6 = 0\) gives \(x=-6\) and \(x - 2=0\) gives \(x = 2\). These are the critical points that divide the number line into intervals.

Step4: Test intervals

The intervals to test are \((-\infty,-6)\), \((-6,2)\) and \((2,\infty)\).

• For \(x<-6\) (e.g., \(x=-7\)): \((-7 + 6)(-7 - 2)=(-1)\times(-9)=9>0\), so the inequality is not satisfied.
• For \(-6
• For \(x>2\) (e.g., \(x = 3\)): \((3 + 6)(3 - 2)=(9)\times(1)=9>0\), so the inequality is not satisfied.

Also, we check the critical points: when \(x=-6\), \((-6 + 6)(-6 - 2)=0\times(-8)=0\leq0\); when \(x = 2\), \((2+6)(2 - 2)=8\times0 = 0\leq0\).

Answer:

\(-6\leq x\leq2\)