knight247 wrote:Find the range of values of x for which x²+7x+12>0
(A)-4<x<-3
(B)x<-4 or x>-3
(C)x<-4 and x>-3
(D)x<3 or x>4
(E)3<x<4

OA is B

I solved the eqn like a regular quadratic equation until I reach (x+3)(x+4)>0
At which point I figure that either both terms are +ve or both are -ve
x>-3 and x>-4
or
x<-3 and x<-4

I really don't know how to choose the correct extremities from the answers I get. Hope someone can give me a detailed explanation. Thanks

You've done everything perfectly so far! Now you just need to condense each of your inequalities.

For the first, x>-3 and x>-4, you need to write one inequality that represents where BOTH of those are true - so you're basically looking for the most restrictive. In this case, if x is a number bigger than -3, then it will take care of BOTH restrictions that x be bigger than -3 AND bigger than -4. So we can write x>-3.

For the second, x<-3 and x<-4, we again need to find the most restrictive inequality. If x is a number smaller than -3, it DOES NOT ensure that x is also less than -4. BUT, if we go with the alternative and say that x is less than -4, then it does cover the restriction that x also be less than -3. So we can write x<-4.

And now we have the 2 sides of the inequality - the expression will be true whenever x>-3 or when x<-4.

Hope this helps!

whit