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Solving an inequality

ATroelstein

New member
Jun 30, 2012
15
I am trying to solve the following inequality:

x3 <= 7x3 - 24x2 + 6x - 10

I have worked it out as follows:

0 <= 6x3 - 24x2 + 6x - 10

10 <= 6x3 - 24x2 + 6x

10 <= 6x(x2 - 4x + 1)

At this point, I'm not sure how to proceed and I'm not sure if the factoring on the last step was helpful. Any advice on how to proceed would be appreciated. Thank you.
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
I am trying to solve the following inequality:

x3 <= 7x3 - 24x2 + 6x - 10

I have worked it out as follows:

0 <= 6x3 - 24x2 + 6x - 10

10 <= 6x3 - 24x2 + 6x

10 <= 6x(x2 - 4x + 1)

At this point, I'm not sure how to proceed and I'm not sure if the factoring on the last step was helpful. Any advice on how to proceed would be appreciated. Thank you.
I would leave it in the form below and solve the LHS and use the real roots of the cubic equation to get intervals to test on.
$6x^3 - 24x^2+ 6x - 10 \geq 0$
$3x^3 - 12x^2+3x-5 \geq 0$

It doesn't look like this one factors easily so you might have to use technology to find the real root(s)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I am trying to solve the following inequality:

x3 <= 7x3 - 24x2 + 6x - 10

I have worked it out as follows:

0 <= 6x3 - 24x2 + 6x - 10
You are fine to here.

10 <= 6x3 - 24x2 + 6x
There is never any good reason to shift that constant to the other side. Knowing that "xy= A" where A is non-zero doesn't tell you anything.
Instead, not that you can divide both sides by the positive number, 2, to get 0<= 3x3- 12x2+ 3x- 5

Now, the only possible rational roots are 5, -5, 1/3, -1/2, 5/3, and -5/3 and it is easy to that none of those are roots. The only root is irrational. Graphing gives a root just a little larger than -1. And the inequality is true for x less than that number.

10 <= 6x(x2 - 4x + 1)

At this point, I'm not sure how to proceed and I'm not sure if the factoring on the last step was helpful. Any advice on how to proceed would be appreciated. Thank you.