Question

How do you solve the following Quadratic Inequality `x^2+2x-15<0`?

Answer 1

The solution is
`-5 < x < 3`

Explanation:

There are more than one ways to solve this inequality.

Solution `A)`
Since `x^2+2x-15 = (x+5)(x-3)`, we can suggest the following reasoning.
The product of two real numbers can be negative if one of them is positive and another is negative.
Therefore, we have two solutions:
`A_1)` `x+5 > 0` AND `x-3 < 0`
`A_2)` `x+5 < 0` AND `x-3 > 0`

The case `A_1` defines `x` as
`x > -5` AND `x < 3`, which defines an interval for `x`:
`-5 < x < 3`

The case `A_2` defines `x` as
`x < -5` AND `x > 3`, which is impossible.

So, the solution is
`-5 < x < 3`

Solution `B)`
As we know, the graph of the quadratic polynomial on the left is parabola. Since the coefficient at `x^2` is positive, this parabola directs its endpoints upward.
Therefore, the only way it can be negative is in-between its roots, where it's equal to zero.
In other words, the solutions to a inequality is the area between solutions to equality
`x^2+2x-15=0`

Obvious solutions are `x=3` and `x=-5`.
So, the solutions are `-5 < x < 2`