Solving quadratic inequalities algebraically without graphing/common-sensing the equality signs 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.
 A: 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 $\;x\in(-\infty,-7)\cup(7,\infty) \;$
A: The problem arises when you divide by $x+7$ since you can't be certain at this point that it is positive. Essentially you disregard the situation where $x+7$ and $x-7$ are both negative.
The alternative does not work as raising both sides to a certain exponent does not work with inequalities as you do not know that you are doing the same thing to both sides. This is more explained here but with squaring, but the same logic applies to this case also.
There are several ways of doing higher degree inequalities, but I will show what I was taught in high school (please correct me if this is not an algebraic way of doing this).
$$x^2-49>0$$
Let's look at the corresponding equation.
$$x^2-49=0$$
$$x=\pm 7$$
Now, we know that these won't work as the original one isn't $x^2-49\geq 0$. If you don't want to do this common-sense as in your title, you can just plug these results back into the inequality.
Now, knowing that these are the only zeroes of the function $f(x)=x^2-49$, we can check the values on the either "sides" of these zeroes and know it will never be zero again. Thus it can't change the sign. This is provable, of course, but this might not be the way you wanted to hear and in that case I will remove this answer.
So, $x=0$ doesn't fulfill the inequality, so $x\notin [-7, 7]$.
$x=100$ and $x=-100$ fulfill the inequality, so $x<-7\lor x>7$
A: If you want a purely algebraic solution then note that the inequality $ab>0$ is only satisfied when $a>0$ and $b>0$, or when $a<0$ and $b<0$. So in your example,
$$
(x+7)(x-7) > 0 \, ,
$$
we need both $x+7$ and $x-7$ to be greater than $0$ (which gives us the solution $x>7$), or we need $x+7$ and $x-7$ to both be less than $0$ (which gives us the solution $x<-7$). However, I think you are much less prone to error if you solve this geometrically.
