Question 1200712
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Answer: <font color=red size=4>(-∞, -3) U (1/3, ∞)</font>


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Explanation:


Let's get everything to one side
3(x^2-1) > -8x
3x^2-3 > -8x
3x^2-3 + 8x > 0
3x^2 + 8x - 3 > 0


To solve that inequality, we'll consider the corresponding equation
3x^2 + 8x - 3 = 0
We need to find the roots or x intercepts.


Factoring may or may not be possible. 
The trial-and-error factoring approach is something I'm not fond of, so I prefer the quadratic formula instead.
Plug in:
a = 3
b = 8
c = -3
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-8+-sqrt((8)^2-4(3)(-3)))/(2(3))}}}


{{{x = (-8+-sqrt(64+36))/(6)}}}


{{{x = (-8+-sqrt(100))/(6)}}}


{{{x = (-8+-  10)/(6)}}}


{{{x = (-8+10)/(6)}}} or {{{x = (-8-10)/(6)}}}


{{{x = (2)/(6)}}} or  {{{x = (-18)/(6)}}}


{{{x = 1/3}}} or  {{{x = -3}}}
Because each root is rational, it turns out that we could have factored previously.
The x = 1/3 leads to 3x = 1 and further leads to 3x-1 = 0
The x = -3 leads to x+3 = 0
We have the factors (3x-1) and (x+3)
Therefore, 3x^2 + 8x - 3 = (3x-1)(x+3)
You can use the FOIL rule on (3x-1)(x+3) to get 3x^2 + 8x - 3 again.


Anyways, the roots we found were:
x = -3
x = 1/3


Draw out a number line. 
Mark -3 on it, and also mark 1/3
We'll label 3 regions
Region A = stuff to the left of -3
Region B = stuff between the number -3 and the number 1/3
Region C = stuff to the right of 1/3
{{{
drawing(600, 200, -7,4,-5,5,
line(-6-3,0,3+3,0),
line(-6,0.2,-6,-0.2),
line(-5,0.2,-5,-0.2),
line(-4,0.2,-4,-0.2),
line(-3,0.2,-3,-0.2),
line(-2,0.2,-2,-0.2),
line(-1,0.2,-1,-0.2),
line(0,0.2,0,-0.2),
line(1,0.2,1,-0.2),
line(2,0.2,2,-0.2),
line(3,0.2,3,-0.2),
locate(-6.2,-0.2,"-6"),
locate(-5.2,-0.2,"-5"),
locate(-4.2,-0.2,"-4"),
locate(-3.2,-0.2,"-3"),
locate(-2.2,-0.2,"-2"),
locate(-1.2,-0.2,"-1"),
locate(-0.2+0.092,-0.2-0.05,"0"),
locate(0.8+0.092,-0.2-0.05,"1"),
locate(1.8+0.092,-0.2-0.05,"2"),
locate(2.8+0.092,-0.2-0.05,"3"),

red(circle(-3,0,0.04)),
red(circle(-3,0,0.06)),
red(circle(-3,0,0.08)),
red(circle(-3,0,0.10)),
red(circle(-3,0,0.12)),

red(circle(1/3,0,0.04)),
red(circle(1/3,0,0.06)),
red(circle(1/3,0,0.08)),
red(circle(1/3,0,0.10)),
red(circle(1/3,0,0.12)),

red(locate(-3,0+1.05,"-3")),
red(locate(1/3,0+1.05,"1/3")),

blue(locate(-5,0+1.25,"A")),
blue(locate((-3+1/3)/2,0+1.25,"B")),
blue(locate(2,0+1.25,"C"))
)

}}}


Pick a value from region A to test.
I'll go for x = -4
3(x^2-1) > -8x
3((-4)^2-1) > -8(-4)
3(16-1) > 32
3(15) > 32
45 > 32
The final result is a true statement.
Therefore, 3(x^2-1)>-8x is true when x = -4
Furthermore, 3(x^2-1)>-8x is true for any value in region A.
We can write x < -3 as part of the solution set


Now let's test region B.
I'll use x = 0
3(x^2-1) > -8x
3(0^2-1) > -8*0
3(0-1) > 0
3(-1) > 0
-3 > 0
That is false, so region B is crossed off the list.


Lastly we need to test region C.
I'll pick x = 2.
3(x^2-1) > -8x
3(2^2-1) > -8*2
3(4-1) > -16
3(3) > -16
9 > -16
This is true, which makes x > 1/3 also part of the solution set.


Our solution set consists of x values such that
x < -3 or x > 1/3


We can rewrite x < -3 as -∞ < x < -3
We can rewrite x > 1/3 as 1/3 < x < ∞
Both of these help us get toward interval notation.


-∞ < x < -3 in interval notation is (-∞, -3)
1/3 < x < ∞ in interval notation is (1/3, ∞)


Those disjoint intervals are glued together with the union operator to arrive at the <font color=red>final answer of (-∞, -3) U (1/3, ∞)</font>


This is what the solution set looks like when graphed on a number line:
{{{
drawing(600, 200, -7,4,-5,5,
line(-6-3,0,3+3,0),
line(-6,0.2,-6,-0.2),
line(-5,0.2,-5,-0.2),
line(-4,0.2,-4,-0.2),

line(-2,0.2,-2,-0.2),
line(-1,0.2,-1,-0.2),
line(0,0.2,0,-0.2),
line(1,0.2,1,-0.2),
line(2,0.2,2,-0.2),
line(3,0.2,3,-0.2),
locate(-6.2,-0.2-0.05,"-6"),
locate(-5.2,-0.2-0.05,"-5"),
locate(-4.2,-0.2-0.05,"-4"),
locate(-3.2,-0.2-0.05,"-3"),
locate(-2.2,-0.2-0.05,"-2"),
locate(-1.2,-0.2-0.05,"-1"),
locate(-0.2+0.092,-0.2-0.05,"0"),
locate(0.8+0.092,-0.2-0.05,"1"),
locate(1.8+0.092,-0.2-0.05,"2"),
locate(2.8+0.092,-0.2-0.05,"3"),

red(circle(-3,0,0.10)),
red(circle(-3,0,0.12)),
red(circle(1/3,0,0.10)),
red(circle(1/3,0,0.12)),

locate(1/3,0+1.05,"1/3"),

red(locate(-5,0+1.25,"x < -3")),
red(locate(2,0+1.25,"x>1/3")),


red(line(1/3+0.12,-0.1,6,-0.1)),
red(line(1/3+0.12,-0.05,6,-0.05)),
red(line(1/3+0.12,0,6,0)),
red(line(1/3+0.12,0.05,6,0.05)),
red(line(1/3+0.12,0.1,6,0.1)),

red(line(4.25-0,0,3.5-0,1)),
red(line(4.25-0.02,0,3.5-0.02,1)),
red(line(4.25-0.04,0,3.5-0.04,1)),
red(line(4.25-0.06,0,3.5-0.06,1)),
red(line(4.25-0.08,0,3.5-0.08,1)),
red(line(4.25-0.1,0,3.5-0.1,1)),

red(line(4.25-0,0,3.5-0,-1)),
red(line(4.25-0.02,0,3.5-0.02,-1)),
red(line(4.25-0.04,0,3.5-0.04,-1)),
red(line(4.25-0.06,0,3.5-0.06,-1)),
red(line(4.25-0.08,0,3.5-0.08,-1)),
red(line(4.25-0.1,0,3.5-0.1,-1)),


red(line(-3-0.12,-0.1,-7,-0.1)),
red(line(-3-0.12,-0.05,-7,-0.05)),
red(line(-3-0.12,0,-7,0)),
red(line(-3-0.12,0.05,-7,0.05)),
red(line(-3-0.12,0.1,-7,0.1)),

red(line(-7+0,0,-6.25+0,1)),
red(line(-7+0.02,0,-6.25+0.02,1)),
red(line(-7+0.04,0,-6.25+0.04,1)),
red(line(-7+0.06,0,-6.25+0.06,1)),
red(line(-7+0.08,0,-6.25+0.08,1)),
red(line(-7+0.1,0,-6.25+0.1,1)),

red(line(-7+0,0,-6.25+0,-1)),
red(line(-7+0.02,0,-6.25+0.02,-1)),
red(line(-7+0.04,0,-6.25+0.04,-1)),
red(line(-7+0.06,0,-6.25+0.06,-1)),
red(line(-7+0.08,0,-6.25+0.08,-1)),
red(line(-7+0.1,0,-6.25+0.1,-1)),

red(locate(-5.5,3,matrix(1,5,"(",-infinity,",",-3,")") )),
red(locate(-5,2,"aka") ),

red(locate(1.5,3,matrix(1,5,"(","1/3",",",infinity,")") )),
red(locate(2,2,"aka") )
)

}}}
Take note of the open holes at -3 and at 1/3.
Verbally we can describe the graph as having "open holes at -3 and at 1/3, with shading everywhere but between those open holes".


An alternative graph to plot is y = 3x^2+8x-3
{{{
drawing(400,400,-6,3,-10,5,
graph(400,400,-6,3,-10,5,-1000,3x^2 + 8x - 3),
circle(-3,0,0.06),
circle(-3,0,0.08),
circle(-3,0,0.10),
circle(-3,0,0.12),
circle(-3,0,0.14),

circle(1/3,0,0.06),
circle(1/3,0,0.08),
circle(1/3,0,0.10),
circle(1/3,0,0.12),
circle(1/3,0,0.14),

locate(-3,0+0.75,"(-3,0)"),
locate(1/3,0+0.75,"(1/3,0)")
)
}}}

The parabola is above the x axis when either x < -3 or when x > 1/3.
In other words, the parabola is on or below the x axis when {{{-3 <= x <= 1/3}}}. Otherwise, it is above the x axis.


Desmos and GeoGebra are two graphing options I recommend.
The graphing option allows a person to quickly arrive at the solution set.
However, I recommend following an algebraic approach and then use a graph to verify (rather than solely rely on a graph to do all the work for you).


A similar problem is found <a href = "https://www.algebra.com/algebra/homework/playground/test.faq.question.1200141.html">here</a>


See <a href = "https://www.mathsisfun.com/algebra/inequality-quadratic-solving.html">this article</a> for further reading.
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