Question

select the values that are solutions to the inequality $x^2 + 3x - 4 > 0$.
□ -6
□ -2
□ 0
□ 5
done

Explanation:

Step1: Factor the quadratic

Factor \(x^2 + 3x - 4\). We need two numbers that multiply to \(-4\) and add to \(3\). Those numbers are \(4\) and \(-1\). So, \(x^2 + 3x - 4=(x + 4)(x - 1)\).

Step2: Find critical points

Set each factor equal to zero: \(x + 4 = 0\) gives \(x=-4\); \(x - 1 = 0\) gives \(x = 1\). These critical points divide the number line into three intervals: \((-\infty, -4)\), \((-4, 1)\), and \((1, \infty)\).

Step3: Test intervals

• For \((-\infty, -4)\), let's pick \(x=-6\). Substitute into \((x + 4)(x - 1)\): \((-6 + 4)(-6 - 1)=(-2)(-7)=14>0\). So, \(x=-6\) is a solution.
• For \((-4, 1)\), let's pick \(x=-2\). Substitute: \((-2 + 4)(-2 - 1)=(2)(-3)=-6<0\). So, \(x=-2\) is not a solution.
• For \(x = 0\) (in \((-4, 1)\)), we already saw this interval is negative, so \(0\) is not a solution.
• For \((1, \infty)\), let's pick \(x = 5\). Substitute: \((5 + 4)(5 - 1)=(9)(4)=36>0\). So, \(x = 5\) is a solution.

Answer:

-6, 5