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

2 Answers
Jul 27, 2017

Answer:

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]}

Jul 27, 2017

Answer:

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.