# How do you solve 1/2x^2+3x<=-6 by algebraically?

Jul 27, 2017

The solution is $S = \emptyset$

#### Explanation:

Let's rewrite the inequality

$\frac{1}{2} {x}^{2} + 3 x \le - 6$

${x}^{2} + 6 x \le - 12$

${x}^{2} + 6 x + 12 \le 0$

Let $f \left(x\right) = {x}^{2} + 6 x + 12$

To calculate the roots of $f \left(x\right)$, we start by calculating the discriminant

$\Delta = {b}^{2} - 4 a c = {6}^{2} - 4 \cdot 1 \cdot 12 = 30 - 48 = - 12$

As,

$\Delta < 0$, there are no roots in $\mathbb{R}$, there are roots in $\mathbb{C}$

So,

$\forall x \in \mathbb{R}$, $f \left(x\right) > 0$

graph{1/2x^2+3x+6 [-18.7, 13.34, -2.56, 13.46]}

Jul 27, 2017

No value for $x$ satisfies:
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} {x}^{2} + 3 x \le - 12$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} {x}^{2} + 3 x \le - 6$

Note that you can multiply both sides of an inequality by any positive value and still maintain the validity and orientation of the inequality.
Multiplying both sides by $+ 2$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 6 x \le - 12$

Note  we can add any amount to both sides of an inequality without effecting the validity or orientation of the inequality.
Adding $12$ to both sides:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 6 x + 12 \le 0$

Note  since ${x}^{2}$ does not have a negative coefficient, it is a parabola which opens upward (i.e. its vertex gives a minimum value for the expression).

Rewriting ${x}^{2} + 6 x + 12$ in "vertex form"
$\textcolor{w h i t e}{\text{XXX}} = {x}^{2} + 6 x + {3}^{2} + 12 - {3}^{2}$

$\textcolor{w h i t e}{\text{XXX}} = {\left(x + 3\right)}^{2} + 3$

$\textcolor{w h i t e}{\text{XXX}} = {\left(x - \left(- 3\right)\right)}^{2} + 3$
with vertex at $\left(x , y\right) = \left(- 3 , 3\right)$

Evaluating $y = \frac{1}{2} {x}^{2} + 3 x$ at this minimum point $\left(x = - 3\right)$
we find
$\textcolor{w h i t e}{\text{XXX}}$minimum value is $\frac{1}{2} \cdot 9 + 3 \cdot \left(- 3\right) = \frac{9}{2} - 9 = + \frac{9}{2}$

Since this value is greater than $- 12$
no value exists for $x$ which satisfies the given inequality.