Question

How do you solve `x^{2} + 3x + 6< 0`?

Answer 1

There are no Real values of `x` satisfying:

`x^2+3x+6 < 0`

In other words, the solution set is empty.

Explanation:

Method 1

Complete the square to find:

`x^2+3x+6 = (x^2+3x+9/4)-9/4+6`

`color(white)(x^2+3x+6) = (x+3/2)^2+15/4`

Note that for any Real value of `x`, we have:

`(x+3/2)^2 >= 0`

and hence:

`x^2+3x+6 = (x+3/2)^2 +15/4 >= 15/4`

So there is no Real value of `x` for which:

`x^2+3x+6 < 0`

Method 2

`x^2+3x+6` is in the form `ax^2+bx+c` with `a=1`, `b=3` and `c=6`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 3^2-4(1)(6) = 9-24 = -15`

Since `Delta < 0`, we can deduce that `x^2+3x+6` has no Real zeros.

Then since the coefficient of `x^2` is positive, we can deduce that:

`x^2+3x+6 > 0`

for all Real values of `x`.

Hence there are no Real solutions to `x^2+3x+6 < 0`