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

Sep 22, 2016

There are no Real values of $x$ satisfying:

${x}^{2} + 3 x + 6 < 0$

In other words, the solution set is empty.

#### Explanation:

Method 1

Complete the square to find:

${x}^{2} + 3 x + 6 = \left({x}^{2} + 3 x + \frac{9}{4}\right) - \frac{9}{4} + 6$

$\textcolor{w h i t e}{{x}^{2} + 3 x + 6} = {\left(x + \frac{3}{2}\right)}^{2} + \frac{15}{4}$

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

${\left(x + \frac{3}{2}\right)}^{2} \ge 0$

and hence:

${x}^{2} + 3 x + 6 = {\left(x + \frac{3}{2}\right)}^{2} + \frac{15}{4} \ge \frac{15}{4}$

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

${x}^{2} + 3 x + 6 < 0$

Method 2

${x}^{2} + 3 x + 6$ is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 3$ and $c = 6$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {3}^{2} - 4 \left(1\right) \left(6\right) = 9 - 24 = - 15$

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

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

${x}^{2} + 3 x + 6 > 0$

for all Real values of $x$.

Hence there are no Real solutions to ${x}^{2} + 3 x + 6 < 0$