Inequality involving conjugate numerator/denominator pairs

Question

Question is to solve:

$$\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)} > 1$$

I thought I could negate terms to make them equal (i.e. $-(x-2)$), but that does not happen. I could subtract $1$ from each side but that would be $(very)^2$ heavy work. Is there any simple way of doing this ?

Answer 1

Here's the first way to solve this question.

The fraction would be greater than $1$ in two cases:

Case I: $(x-2)(x-4)(x-7)$ and $(x+2)(x+4)(x+7)$ are both positive and $(x-2)(x-4)(x-7) > (x+2)(x+4)(x+7)$.

Case II: $(x- 2)(x-4)(x-7)$ and $(x+2)(x+4)(x+7)$ are both negative and $(x-2)(x-4)(x-7) < (x+2)(x+4)(x+7)$.

---

Another way, and the preferable one, is to take the denominator to the other side, but here again, consider cases. When the denominator is negative, then change the direction of the inequality and when it's not, let it stay the same.

Do not multiply it straightaway because the denominator has different signs in different intervals.

Focus on the "and", which signifies that in each case, you first find the intervals for the first condition, then the interval(s) for the second condition, and then take their intersection.

Answer 2

Looks like you will obtain a quadratic inequality if you multiply by the denominator and then expand.

Consider the following cases:

$x<-7$
$-7<x<-4$
$-4<x<-2$
$x>-2$

The sign of the denominator in each case tells you what to do with the inequality sign when multiplying.

Answer 3

Another way would be to multiply throughout by $(x+2)^2(x+4)^2(x+7)^2$, which is positive ($x =-2,-4,-7$ are not allowable anyway), to get $$(x^2-4)(x^2-16)(x^2-49)>(x+2)^2(x+4)^2(x+7)^2$$

Collecting terms in the LHS and factoring, this is equivalent to $$-2(x+7)(x+4)(x+2)(13x^2+56)>0$$ and now checking intervals where LHS can change signs gives you the answer $x \in (-\infty, -7) \cup (-4,-2)$.