Question

How do you solve the inequality `x^2+3x-28<0`?

Answer 1

`-7 < x < +4`

Explanation:

`x^2+3x - 28 = x^2+7x-4x-48 = x(x+7)-4(x+7) = (x-4)(x+7)`

So, the inequality is

`(x-4)(x+7)<0`

Now. the left hand side vanishes for `x = -7` and `x=+4`.

If `x<-7`, then both the factors `(x+7)` and `(x-4)` are negative, and the product is positive. On the other hand, `if x> +4`, both factors are positive, leading once again to a positive product.

Thus the inequality is only satisfied when

`-7 < x < +4`

where the facor `(x+7)` is positive and `(x-4)` is negative.

You can see this in the graph below

graph{x^2+3x-28 [-10, 10, -40, 40]}