I don't get when you use "or" or "and" in inequalities.
Like for this quadratic inequality:
x^(2) +x -12 > 0
becomes x < -4
x > 3
why is the answer {x|x<-4 or x>3} and not {x|3<x<-4} ?
Your quadratic inequality has TWO solutions: x < -4 AND x > 3.
The 2nd arrangement where you have x in the center of the inequality is normally used with compound inequalities.
If you look at the graph, it would be clear that the part of the curve which is above the x-axis is in two separate parts, therefore the answer is
x<-4 and x>3.
On the other hand if the question had been x^(2) +x -12 < 0 , then the solution will be continuous on the number line, namely -4<x<3.
See graph:
http://img529.imageshack.us/img529/2064/1285810976.png
When solving quadratic inequalities, we need to find the values of 'x' that satisfy the given inequality. In this case, the inequality is x^2 + x - 12 > 0.
To solve this quadratic inequality, we can find the x-intercepts of the associated quadratic equation by factoring it or using the quadratic formula. The x-intercepts are the values of 'x' for which the quadratic equation equals zero.
The quadratic equation x^2 + x - 12 = 0 factors as (x - 3)(x + 4) = 0. Therefore, the x-intercepts are x = 3 and x = -4.
Now, let's consider the intervals between these x-intercepts on the number line:
- Infinity <----(-4)----(-)-------3------(+)----> Infinity
To determine the sign of the quadratic expression x^2 + x - 12 in each interval, we choose a value in that interval and plug it into the inequality. For example, if we choose x = -5, we get (-5)^2 - 5 - 12 = 0 > 0, so this section is positive. If we choose x = 0, we get (0)^2 + 0 - 12 = -12 < 0, so this section is negative. Finally, if we choose x = 4, we get (4)^2 + 4 - 12 = 16 > 0, so this section is positive.
To satisfy the inequality x^2 + x - 12 > 0, we need the quadratic expression to be positive. Looking at the intervals, we see that the quadratic expression is positive when x is less than -4 OR when x is greater than 3. In interval notation, we write this as:
{x | x < -4 or x > 3}
So, the solution to the inequality x^2 + x - 12 > 0 is {x | x < -4 or x > 3}. This means that any value of 'x' which is less than -4 or greater than 3 will make the inequality true.