I've just been presented an inequality at GCSE (high school) level that, solving algebraically (the method used for linear inequalities), doesn't work correctly which has caught me by surprise. Research always leads to graphed or common-sensed solutions. Can anyone explain why this doesn't work and if there's an algebraic manipulation that leads to the right answers?
$$x^2 – 49 > 0$$
Using the quadratic identity
$$(x + 7)(x – 7) > 0$$
Divide both sides by (x+7)…
$$x – 7 > 0$$
$$x > 7$$
Divide both sides by (x-7)…
$$x + 7 > 0$$
$$x > -7$$
Alternatively,
$$x^2 – 49 > 0$$
$$x^2 > 49$$
$$x > \pm7$$
The negative solution has to be $$x < -7$$ but I don't know how that's arrived at through algebra.
Best Answer
It is much fun and pretty enlightening to do this kind of exercises with a ping-pong between algebra and geometry. Anyway, if you want to avoid that , then we have the following algebraic facts which, funny enough, are pretty easy to prove precisely by means of geometry...but not only, as they follow from the definition of absolute value:
$$\text{For}\;0<a\in\Bbb R\;,\;\;\begin{cases}|x|\le a\iff -a\le x\le a\\{}\\ |x|>a\iff x<-a\;\;\text{or}\;\;x>a\end{cases}$$
Thus, in your case, we could do as follows (all algebra, not geometry):
$$x^2-49>0\implies x^2>49\stackrel{\text{taking square roots}}\implies |x|>7\iff x<-7\;\;\text{or}\;\;x>7$$
and in intervals notation, the solution is $\;x\in(-\infty,-7)\cup(7,\infty) \;$