# Thread: [SOLVED] Show that 2 + 8x - 2x^3 = 0 has three real solutions

1. ## [SOLVED] Show that 2 + 8x - 2x^3 = 0 has three real solutions

Hi

How do I show that 2 + 8x - 2x^3 = 0 has three real solutions?

I know you can say it has at most three solutions from the fundamental theorem of algebra, and I presume the easiest way to show the three solutions are real and not complex is to actually find them... But how ought one to go about this?
I can't see any convenient factoring up the equation..

Moderator edit: This thread has been moved from the Pre-Calculus to the Calculus subforum. This comment is made so that some of the posts (which I choose not to delete) in this thread make sense. For the record, if a question requires differentiation or integration then it is clearly a calculus question and belongs in the Calculus subforum.

2. Originally Posted by PTL
Hi

How do I show that 2 + 8x - 2x^3 = 0 has three real solutions?

I know you can say it has at most three solutions from the fundamental theorem of algebra, and I presume the easiest way to show the three solutions are real and not complex is to actually find them... But how ought one to go about this?
I can't see any convenient factoring up the equation..
If there are no specifications on how to do it, I'd just graph it.

3. Originally Posted by PTL
Hi

How do I show that 2 + 8x - 2x^3 = 0 has three real solutions?

I know you can say it has at most three solutions from the fundamental theorem of algebra, and I presume the easiest way to show the three solutions are real and not complex is to actually find them...
You know more than this: any real polynomial of odd degree has at least one real zero.

I can't see any convenient factoring up the equation..
Well, maybe you are allowed to use a little calculus: Determine the high and low point (if any) of the graph of $f(x) :=-2x^3+8x+2$
You will see that there is a high point above the x-axis and a low point below the x-axis. - Therefore, from the intermediate value theorem, we can conclude that there must be three real zeros of f.

4. Originally Posted by Failure
Well, maybe you are allowed to use a little calculus: Determine the high and low point (if any) of the graph of $f(x) :=-2x^3+8x+2$
You will see that there is a high point above the x-axis and a low point below the x-axis. - Therefore, from the intermediate value theorem, we can conclude that there must be three real zeros of f.
Ah, but this is the Pre-calculus forum, so we cannot use this method. Either that, or the OP should have posted this in the Calculus subforum.

So without calculus, you'd best graph the function. If you really want to solve it algebraically, I guess you could try the cubic formula (see here: Cubic function - Wikipedia, the free encyclopedia) but I wouldn't recommend it. It's messy.

01

5. Thank you all.

It was really calculus, I suppose... I just figured it was too basic to put in the calc forum...
I was just reading the notice about posting in the wrong forum... What is pre-calculus exactly, if there's no calculus involved?

6. That's the thing. There is no set definition of what Precalculus is, so I don't fault you for posting in the wrong forum. Where I live:

Precalculus =
advanced algebra (many schools call this "College Algebra") plus
trigonometry (and yes, we have a separate trig subforum) plus
analytic geometry (conic sections, polar coordinates, vectors) plus
a little linear algebra (systems of linear equations and matrices) plus
a little discrete math (binomial theorem, sequences & series, mathematical induction).

01

7. Originally Posted by Failure
You know more than this: any real polynomial of odd degree has at least one real zero.

Well, maybe you are allowed to use a little calculus: Determine the high and low point (if any) of the graph of $f(x) :=-2x^3+8x+2$
You will see that there is a high point above the x-axis and a low point below the x-axis. - Therefore, from the intermediate value theorem, we can conclude that there must be three real zeros of f.
f'(x) = 8 - 8x^2
1 - x^2
This equals 0 when x = 1 or -1.
So there is a max/min at (1,8) and (-1,-7)?

8. Originally Posted by PTL
f'(x) = 8 - 8x^2
1 - x^2
This equals 0 when x = 1 or -1.
So there is a max/min at (1,8) and (-1,-7)?
$f'(x)=8-6x^2=0$

$4-3x^2=0$

$x^2=\frac{4}{3}$

$\Rightarrow\text{ critical values at }x=\pm\frac{2\sqrt{3}}{3}$

9. Originally Posted by VonNemo19
$f'(x)=8-6x^2=0$

$4-3x^2=0$

$x^2=\frac{4}{3}$

$\Rightarrow\text{ critical values at }x=\pm\frac{2\sqrt{3}}{3}$

$6x^2$, of course...

And is the fact that
a) there are three roots
b) the max and min are on either side of the x-axis
enough to assert that the 3 roots are Real?
There can't be a complex root whose real part is between the max and min?

10. Originally Posted by PTL
$6x^2$, of course...

And is the fact that
a) there are three roots
b) the max and min are on either side of the x-axis
enough to assert that the 3 roots are Real?
There can't be a complex root whose real part is between the max and min?
a) Yes. By the intermiadate value theorem. Note how the curve MUST cross the x-axis 3 times.

b) yes they are.

11. Thank you very much.

12. Originally Posted by PTL
Hi

How do I show that 2 + 8x - 2x^3 = 0 has three real solutions?

I know you can say it has at most three solutions from the fundamental theorem of algebra, and I presume the easiest way to show the three solutions are real and not complex is to actually find them... But how ought one to go about this?
I can't see any convenient factoring up the equation..
you can use the Intermediate Value Theorem -- from Wolfram MathWorld

13. Originally Posted by PTL
Hi

How do I show that 2 + 8x - 2x^3 = 0 has three real solutions?

I know you can say it has at most three solutions from the fundamental theorem of algebra, and I presume the easiest way to show the three solutions are real and not complex is to actually find them... But how ought one to go about this?
I can't see any convenient factoring up the equation..
Let $f(x)=2 + 8x - 2x^3.$

$f(-2)\ =\ 2+8(-2)-2(-2)^2\ =\ 2\ >\ 0$

$f(-1)\ =\ 2+8(-1)-2(-1)^2\ =\ -8\ <\ 0$

$f(0)\ =\ 2\ >\ 0$

$f(3)\ =\ 2+8(3)-2(3)^3\ =\ -28\ <\ 0$

Hence, by the intermediate-value theorem, there is a root to the equation $f(x)=0$ in each of the intervals $(-2,\,-1),$ $(-1,\,0)$ and $(0,\,3).$

EDIT: Sorry, I didn’t read the replies before posting.

14. Thanks, a concrete numerical solution is always helpful. Even if theoretical mathematical waffle is all that's needed..