Question

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

Answer 1

The solution is `S=O/`

Explanation:

Let's rewrite the inequality

`1/2x^2+3x<=-6`

`x^2+6x<=-12`

`x^2+6x+12<=0`

Let `f(x)=x^2+6x+12`

To calculate the roots of `f(x)`, we start by calculating the discriminant

`Delta=b^2-4ac=6^2-4*1*12=30-48=-12`

As,

`Delta<0`, there are no roots in `RR`, there are roots in `CC`

So,

`AA x in RR`, `f(x)>0`

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

Answer 2

No value for `x` satisfies:
`color(white)("XXX")1/2x^2+3x <= -12`

Explanation:

Given
`color(white)("XXX")1/2x^2+3x <= -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`
`color(white)("XXX")x^2+6x <=-12`

Note [2] 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:
`color(white)("XXX")x^2+6x+12 <= 0`

Note [3] 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+6x+12` in "vertex form"
`color(white)("XXX")=x^2+6x+3^2+12-3^2`

`color(white)("XXX")=(x+3)^2+3`

`color(white)("XXX")=(x-(-3))^2+3`
with vertex at `(x,y)=(-3,3)`

Evaluating `y=1/2x^2+3x` at this minimum point `(x=-3)`
we find
`color(white)("XXX")`minimum value is `1/2 * 9 +3 *(-3) = 9/2 -9 = +9/2`

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