Algebra 2 Super Challenge Problem!

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November 4th 2012, 11:13 PM
odysseyofnoises

Algebra 2 Super Challenge Problem!

Hello everyone, this is my first post in the Math Help Forum. I was wanting to know if anybody out there can prove that x = 3 (without just 'plugging it' in) in this equation that my friend made:

³√(8x+3) - ³√(x-4) = 4 (third roots)

Here's what I have of the proof so far: View image: Proof of the Super Equation( WIP) Sorry if you can't follow it lol...
What would you recommend I do as the next step? The numbers are really getting out of control...
November 5th 2012, 01:54 AM
BobP

Re: Algebra 2 Super Challenge Problem!

At the point where you have $8x+3 = 64 + .....$ , rearrange the equation as

$7(x-4) - 12(x-4)^{2/3}-48(x-4)^{1/3} - 29 = 0.$

Now make a substitution, $z = (x-4)^{1/3},$ and you arrive at the cubic equation

$7z^{3}-12z^{2}-48z-29 = 0.$

Whether or not you accept trial values for $z$ (which is a normal method) I don't know. ( $z = -1$ is a fairly obvious solution.) The alternative is to use perhaps Cardano's method.
November 5th 2012, 04:46 PM
odysseyofnoises

Re: Algebra 2 Super Challenge Problem!

How did you go from the 8x+3 = 64... to the 7(x-4) one? I don't see how the flow of logic works from one step to the next. I understand the substitution part though. I got this: View image: Proof of the Super Equation( WIP)
November 6th 2012, 12:08 AM
BobP

Re: Algebra 2 Super Challenge Problem!

$-7x+57 = -7x+28+29=-7(x-4)+29.$
November 6th 2012, 05:55 AM
odysseyofnoises

Re: Algebra 2 Super Challenge Problem!

Oh, that makes sense :P you just rewrite those into an (x-4) term so that it can be replaced with substitution. Now I'll just work with the cubic function soon (maybe not today, have too much homework...). And yes, I accept trial values for z, otherwise I would probably use some extremely complicated method...
November 6th 2012, 07:27 PM
odysseyofnoises

Re: Algebra 2 Super Challenge Problem!

Now, I am doing polynomial division between (z+1) and the cubic function. I am getting a remainder, what course of action do you recommend now? here's what I have so far: View image: Proof of the Super Equation( WIP)
November 6th 2012, 07:35 PM
MarkFL

Re: Algebra 2 Super Challenge Problem!

You should not get a remainder in the polynomial long division. Doing the division, I find:

$7z^3-12z^2-48z-29=(z+1)(7z^2-19z-29)$
November 6th 2012, 07:44 PM
odysseyofnoises

Re: Algebra 2 Super Challenge Problem!

I think I see what I did wrong here... bad arithmetic. Oh well, someday I'll learn to count :P
November 6th 2012, 08:16 PM
topsquark

Re: Algebra 2 Super Challenge Problem!

Quote:

Originally Posted by BobP

At the point where you have $8x+3 = 64 + .....$ , rearrange the equation as

$7(x-4) - 12(x-4)^{2/3}-48(x-4)^{1/3} - 29 = 0.$

Now make a substitution, $z = (x-4)^{1/3},$ and you arrive at the cubic equation

$7z^{3}-12z^{2}-48z-29 = 0.$

Whether or not you accept trial values for $z$ (which is a normal method) I don't know. ( $z = -1$ is a fairly obvious solution.) The alternative is to use perhaps Cardano's method.

For the record the cubic has the 3 real solutions as
$z = -1,~\frac{19 \pm \sqrt{1173}}{14}$

So
$x = 3,~\frac{10587 \pm 282\sqrt{1173}}{343}$

All of these are correct according to the original equation, but I would highly recommend the last two solutions be done on a calculator. (You'll see why when you plug them into the original equation.)

Okay, I was bored. (Nerd)

-Dan