Question

How do you solve `14-5x-x^2<0`?

Answer 1

`x<-7 uu x>2`

Explanation:

Just to make things easier, we multiply both sides by `-1` and flip the inequality sign: `x^2+5x-14>0`. This way the coefficient before `x^2` is positive. (Note: this is not a necessary step.)

We factorize the expression: `x^2+5x-14=(x+7)(x-2)>0` to find the roots: `x=-7` or `2`.

These are the only two possible points where the expression can change signs. Thus, there are three regions that have the same sign:

• `x<-7`
• `-7 < x < 2`
• `x>2`

Since there are no repeated roots, the signs of the region alternate. We substitute an arbitrary number in the second region to find the sign: `x=0`. Then `x^2+5x-14=0^2+5*0-14=-14<0`. Thus, the second region is negative.

Therefore, the first and third regions are positive (you can use substitution to be sure). Looking at the inequality, `x^2+5x-14>0`, we notice that we need to find regions where the value is positive.

The solution is the first and third regions: `x<-7 uu x>2`.