SOLUTION: What is the answer to thus quadratic inequality. 3(x^2-1)>-8x. The solution should be written in interval notation.

Question 1200712: What is the answer to thus quadratic inequality. 3(x^2-1)>-8x. The solution should be written in interval notation.
Found 3 solutions by greenestamps, math_tutor2020, Alan3354:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

(1) Rewrite the inequality in standard form with "0" on the right -- just as we would do if we were solving a quadratic equation:

-->

(2) Factor the quadratic expression:

-->

(3) Determine the solution set.

The expression is EQUAL to 0 at x=-3 and x=1/3. The sign of the evaluated expression can only change at those two values of x.

So, because the inequality is a strict inequality (it can't be equal to zero), we have three intervals of interest: (-infinity,-3), (-3,1/3), and (1/3,infinity). There are two basic ways to determine which of those intervals are part of the solution set.

One elementary way is to choose a test value in each of the intervals and find the intervals on which the inequality is satisfied. I will leave that to you.

Another elementary way is to know that the graph of the quadratic expression is an upward-opening parabola, so the expression is negative only between x=-3 and x=1/3. We then know on which intervals the expression is greater than zero.

ANSWER: (-infinity,-3) U (1/3,infinity)

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answer: (-∞, -3) U (1/3, ∞)

======================================================================

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

or

or

or
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

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 final answer of (-∞, -3) U (1/3, ∞)

This is what the solution set looks like when graphed on a number line:

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

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 . 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 here

See this article for further reading.

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
There is no answer because there is no question.
--------
Questions are answered.
Equations and inequalities are solved.
RELATED QUESTIONS
2/3+3/5>=x/15 what is the solution set written in interval... (answered by ewatrrr)
How do I solve the quadratic inequality x^2-8x+12<0 and answer it in interval... (answered by lwsshak3)
Solve the quadratic inequality. Write the solution set in interval notation.... (answered by josgarithmetic)
Solve the quadratic inequality. Write the solution set in interval notation. x^2-3x-10>0 (answered by josgarithmetic)
solve the inequality and write the solution in interval notation x+1/x+3 >... (answered by stanbon)
solve the quadratic inequality. Write the solution set in interval notation.... (answered by Edwin McCravy)
Find the solution for inequality. Write the answer in interval notaion.... (answered by rapaljer)
What is the solution (in interval notation) to the following inequality below? -10... (answered by Theo,bucky)
Using interval notation, the solution to the inequality xsqt+8x-20<0 is all in the... (answered by Fombitz)