Question

solve the given inequality; then graph the solution set.

1. $x^{2}+2x-3 > 0$

Explanation:

Step1: Factor the quadratic

$x^2 + 2x - 3 = (x+3)(x-1)$

Step2: Find critical points

Set $(x+3)(x-1)=0$
Solutions: $x=-3$, $x=1$

Step3: Test interval $(-\infty, -3)$

Substitute $x=-4$: $(-4+3)(-4-1)=(-1)(-5)=5>0$, so interval is valid.

Step4: Test interval $(-3, 1)$

Substitute $x=0$: $(0+3)(0-1)=(3)(-1)=-3<0$, so interval is invalid.

Step5: Test interval $(1, \infty)$

Substitute $x=2$: $(2+3)(2-1)=(5)(1)=5>0$, so interval is valid.

Step6: Graph the solution set

Open circles at $x=-3$ and $x=1$, shade regions left of $-3$ and right of $1$.

Answer:

The solution set is $(-\infty, -3) \cup (1, \infty)$.
For the graph: Draw open dots at $x=-3$ and $x=1$ on the number line, shade all values to the left of $-3$ and all values to the right of $1$.