Question 57480: 1) Solve x2-2x-3>0.
a. (-3, 1)
b. (-1, 3)
c. (- ∞ , -3) U (1, ∞ )
d. (- ∞ , -1) U (3, ∞ )
Found 2 solutions by funmath, stanbon:
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! 1) Solve  .
a. (-3, 1)
b. (-1, 3)
c. (- infinity , -3) U (1, infinity )
 . (- infinity , -1) U (3, infinity )
:
Here's why:
Find the critical values that make this =0. Factor and solve.
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!!!
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Solve f(x)=x^2-2x-3>0.
Consider the equality:
f(x)=x^2-2x-3=0
Factor to get:
(x-3)(x+1)=0
x=3 or x=-1
Draw a number line.
Put -1 and 3 iin their appropriate positions.
The break the line into three intervals:
I.: (-inf,-1)
II: (-1,3)
III: (3,inf)
Check a point in each interval to determine where the
solutions of the inequal lie:
I: say x=-10, then f(-10)>0 so I is part of the solution set.
II: say x=0, then f(0)=-3<0 so II is not part of the soution.
III: say x=10, then f(10)>0 so III is part of the solution set.
Conclusion: The solution of the inequality is (-inf,-1) or (3,inf).
Cheers,
Stan H.
|
|
|
