Question 57480
1) Solve {{{x^2-2x-3>0}}}.
a. (-3, 1)
b. (-1, 3)
c. (- infinity , -3) U (1, infinity )
{{{highlight(d)}}}. (- infinity , -1) U (3, infinity )
:
Here's why:
 {{{x^2-2x-3>0}}}  Find the critical values that make this =0.  Factor and solve.
{{{(x+1)(x-3)=0}}}
x+1=0    and x-3=0 
x+1-1=0-1 and x-3+3=0+3
x=-1  and x=3
Test the interval (-infinty,-1)
Test -10 (-10+1)(-10-3) a (-)(-) is a positive.  This interval is>0
Test the interval (-1,3)
Test 0 (0+1)(0-3) a (+)(-) is a negative.  This interval is not >0
Test the interval (3,infinfty)
Test 10 (10+1)(10-3) a (+)(+) is a positive. This interval is>0
:
Therefore the solution is (-infinity,-1)U(3,infinity)
Happy Cacluating!!!