x^2 + 2x > 3 -- inequality question

Guitarman

New member
Joined
Aug 22, 2006
Messages
9
I'm trying my best to understand these type of questions but I am having some trouble. Here's what I've got:

1) x^2 + 2x > 3

x^2 + 2x - 3 > 0
(x - 1)(x + 3) > 0
1 > x > - 3 => interval: [1,-3] <---solution???

Also I have another one:

2) 3 - |2x + 4| < 1

I'm not sure how to go about this one. Thank you for any help.
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
1) Solve the related equality, "x<sup>2</sup> + 2x - 3 = 0", to find the endpoints of the intervals. Then use whatever method you've learned to test the intervals. You want the intervals on which x<sup>2</sup> + 2x - 3 is positive, not negative. (Hint: Look at the graph of x<sup>2</sup> + 2x - 3. Where is it above the x-axis?)

And since this is a strict inequality (that is, no "or equal to"), the endpoints (where the quadratic is equal to zero) cannot be part of the solution.

2) You can add and subtract, giving you:

. . . . .2 < |2x + 4|

Then split this into the two linear inequalities that this absolute-value inequality generates, and solve each.

For instance:

. . . . .1 < |x - 5|

. . . . .|x - 5| > 1

. . . . .x - 5 > 1, so x > 6
. . . . .x - 5 < -1, so x < 4

Follow the same procedure with your exercise.

Eliz.
 
Top