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

Sep 11, 2015

The solution is
$- 5 < x < 3$

#### Explanation:

There are more than one ways to solve this inequality.

Solution A)
Since ${x}^{2} + 2 x - 15 = \left(x + 5\right) \left(x - 3\right)$, 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
${x}^{2} + 2 x - 15 = 0$

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