My question is just making sure that my working is correct and that I understand properly (self teaching, can get confused...)
So question :
Find the set of values for which
$$x^2 -4x-12 < 0$$
I can factor this to
$$(x-2)(x+6) < 0$$
So this means that $x = 2$ or $x = -6$
This then get expressed as
$$-2<x<6$$
I can't really explain why though...
Is this because $x - 2 = 0$ so $x > -2$ ...?
If I create a graph of $y = (x-2)(x+6)$ the range of values between the two x intercepts are $-6 $ to $2$....
So from that I would expect the equalities to be
$$-6<x<2$$
not
$$-2<x<6$$
I'm not sure what I'm missing here?
Thanks
$$\;x^2-4x-12=(x+2)(x-6)$$
is a parabola that opens up (a "smiling" one) or a convex upwards one, and it vanishes at $\;x=-2,6\;$ . If you make even a raw drawing of this, you will clearly see that
$$x^2-4x-12<0\iff -2<x<6$$
The above's already a complete, mathematical formal proof if you've already studied quadratic functions and parabolas from a geometric/algebraic point of view.
One can also prove the inequality in a purely algebraic way:
$$(x+2)(x-6)<0\iff \begin{cases}x+2>0\;\;\text{and}\;\;x-6<0\\or\\x+2<0\;\;\text{and}\;\;x-6>0\end{cases}$$
The second possibility above is impossible as this would imply $\;x>-2\;\;\wedge\;\;x>6\;$ , and there's no such a thing, so it must be the first one, which is the same we got in the first part above.