Solve for x, x being a member of real numbers (2 q.)

[Guest]

k I think I need a little review on these, because I keep writing down the symbols wrong. I know how to get the numbers, but I seem to mess up on the symbols.

a) (x-2)(x+4)<0
ans.= -4<x<2
do you just change the dirrection of the symbol based on common sense? cuz it has to be greater than -4, and less that 2, it can't be the other way, or it wouldn't make any sense?

b) X^2+x-2 greater than or equal to 0
so you would first factor: (x+2)(x-1) greater than or equal to 0
ans: = x less than or equal to -2 or x greater than or equal to 1

Now why is there an OR? and now like x in the middle with signs on either side?

EEEEp Help please

 

[stapel]

a) This is a quadratic, a positive one, right? One that graphs as an upward-opening parabola, right? The zeroes are clearly at x = -4 and x = 2. These zeroes divide the number line into the intervals (-infinity, -4), (-4, 2), and (2, +infinity). On which of these intervals will the parabola be below the x-axis?

b) You ask why there is an "or". Since the original inequality was "greater than or equal to", then the solution intervals will also be "or equal to".

Eliz.

 

[Guest]

mm I still don't really get how to do this (I wasn't there for the lesson) but for these types of questions there is a chart made and a graph. like for b) the chart headings are x<0, 0<x<2, 2<x<3, and x>3 then under these headins are -ve and +ve's . under )<x<2 there is a +,-,- therefore equalling to a '+' and for x>3 +,+,+ is under this heading, therefore equalling to a positive as well.but how do they get the -ves and +ves?