Question

For how many integers `a` is it true that `a^2-8` is a negative number?

Answer 1

Possible integer values of `a`, for which `a^2-8<0` are `-2,-1,0,1` and `2`.

Explanation:

Well for `a^2-8` to be a negative number, it tantamounts to solve the inequality `a^2-8<0`.

Now this is nothing but `(a-2sqrt2)(a+2sqrt2)<0`

Note that

1. for `a<-2sqrt2`, both `a-2sqrt2` and `a+2sqrt2` are negative and hence `a^2-8>0`
2. for `a>2sqrt2`, both `a-2sqrt2` and `a+2sqrt2` are positive and hence `a^2-8>0`
3. But for `-2sqrt2 < a < 2sqrt2`, while `a-2sqrt2` is negative, `a+2sqrt2` is positive and hence `a^2-8<0`

Hence possible integer values of `a`, for which `a^2-8<0` are `-2,-1,0,1` and `2`.