Algebra and Cube Roots...

Discussion in 'Community' started by realityisterror, Jan 12, 2005.

1. realityisterror macrumors 65816

Joined:
Aug 30, 2003
Location:
Snellville, GA
#1
Alright... you math geniuses: some help with an extra credit problem that's been bugging me and my dad...

cuberoot(x+9) - cuberoot(x-9) = 3

The thing is, I know the answer is sqrt(80) through guess and check , but I want to know how to back it up mathematically. A nudge in the right direction of how to deal with it would be much appreciated.

reality

Staff Member

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Feb 5, 2004
Location:
New Zealand
#2
I started by cubing each side. Although this initially gives a rather ugly equation, you can simplify it a bit and substitute in the original equation, which gives you a much easier equation to solve. Ask me if you want the full solution or some further hints.

3. realityisterror thread starter macrumors 65816

Joined:
Aug 30, 2003
Location:
Snellville, GA
#3
Explain cubing both sides... I think I know what you mean, but I'm not sure...
You don't mean (I hope): (x+9)-(x-9)=27
So on the left side, I follow the pattern A^3 + B^3 + 2A^2B + 2AB^2??
BTW, did you end up getting sqrt(80), or did you not work it out completely?

I guess I need a bit bigger of a nudge...

Thanks,
reality

Staff Member

Joined:
Feb 5, 2004
Location:
New Zealand
#4
Close, it's actually A^3 + 3A^2B + 3AB^2 + B^3 for (A+B)^3, or A^3 - 3A^2B + 3AB^2 - B^3 for (A-B)^3 (which should be used in this case). With A = (x+9)^(1/3) and B=(x-9)^(1/3), this comes out as:
(x+9) - 3(x+9)^(2/3)*(x-9)^(1/3) + 3(x+9)^(1/3)*(x-9)^(2/3) - (x-9) = 27

Yeah, I got that, although -80^0.5 is also valid.

5. mlw1235 macrumors 6502

Joined:
Jul 16, 2004
Location:
Milwaukee, WI
#5

Thanks alot, now it's buggin me also!

btw, What level math class is this for? And do you know lots of factoring out stuff (as in fractional exponents)?

6. realityisterror thread starter macrumors 65816

Joined:
Aug 30, 2003
Location:
Snellville, GA
#6
you rule hexmonkey.. it still took me a bit of work, but i got it figured out now...
i'm in 9th grade geometry, but the problem is 10th grade algebra 2 level. it's fairly complicated to type out the whole problem on the computer, but maybe i'll find a math text editor and type it out tomorrow...

yawn.. should have been in bed a while ago...

thanks,
reality

7. jsw Moderator emeritus

Joined:
Mar 16, 2004
Location:
Andover, MA
#7
If anyone's curious about the solution, now that realityisterror has finished, I'll provide it here. Why? Because, after vast years of math during school, my current job requires none, and this was the first time I've worked my brain with a problem of this type in well over a decade. I coincidentally used HexMonkey's approach as well (cubing both sides), and reached the same intermediate results.

Further reducing HexMonkey's intermediate results (above) yields:

(x+9)^(1/3) * (x-9)^(2/3) - (x+9)^(2/3) * (x-9)^(1/3) = 3

Substituting

[(x-9)^(1/3) + 3] for (x+9)^(1/3) in the first part of the above equation,

and

[(x+9)^(1/3) - 3] for (x-9)^(1/3) in the second part

yields, after reduction:

(x+9)^(2/3) + (x-9)^(2/3) - 7 = 0

If we let a = (x+9)^(1/3) and b = (x-9)^(1/3), then the above becomes:

a^2 + b^2 - 7 = 0

Also, from the initial equation (first post), we have:

a - b = 3

hence

b = a - 3

and we get

a^2 + (a-3)^2 - 7 = 0

which reduces to

a^2 - 3a + 1 = 0

therefore

a = (3 +/- sqrt(5))/2

hence, if we assign z = (3 +/- sqrt(5)),

x = (z^3)/8 - 9

If we use z = 3 + sqrt(5), then

z^3 = 72 + 32*sqrt(5),

meaning

x = 9 + 4* sqrt(5) - 9 = 4 * sqrt(5) = sqrt(16) * sqrt(5) = sqrt(80)

if we use z = 3 - sqrt(5), then we get, similar to above, x = -sqrt(80)

That was fun.

Staff Member

Joined:
Feb 5, 2004
Location:
New Zealand
#8
Here's my solution, which solves it slightly differently from jsw's.

Cube each side, giving us
(x+9) - 3(x+9)^(2/3)*(x-9)^(1/3) + 3(x+9)^(1/3)*(x-9)^(2/3) - (x-9) = 27

Which simplifies to
-3(x+9)^(2/3)*(x-9)^(1/3) + 3(x+9)^(1/3)*(x-9)^(2/3) = 9
-(x+9)^(2/3)*(x-9)^(1/3) + (x+9)^(1/3)*(x-9)^(2/3) = 3

Factorising, we get
(x+9)^(1/3)*(x-9)^(1/3)*(-(x+9)^(1/3) + (x-9)^(1/3)) = 3
-(x+9)^(1/3)*(x-9)^(1/3)*((x+9)^(1/3) - (x-9)^(1/3)) = 3

Substite (x+9)^(1/3) - (x-9)^(1/3) for 3 (from the original equation), giving
-3(x+9)^(1/3)*(x-9)^(1/3) = 3
(x+9)^(1/3)*(x-9)^(1/3) = -1

Cube each side again and simplify
(x+9)(x-9) = -1
x^2 - 81 = -1
x^2 = 80
x = 80^0.5 or -80^0.5

9. jsw Moderator emeritus

Joined:
Mar 16, 2004
Location:
Andover, MA
#9
Simpler, that.

Apparently, those years of inactivity have left my mental mathematical muscles a wee bit out of shape!

Joined:
Oct 26, 2004
Location:
#10
huh?

nope i dont even understand a bit about what you lot have said
and im meant to be one of the top GCSE maths pupils in the UK...

i tried to cube the entire equation so i got:

cuberoot(x+9) - cuberoot(x-9) = 3

x+9-(x-9)=27???
that means
x+9-x+9=27
18=27

11. 4409723 Suspended

Joined:
Jun 22, 2001
#11
Or if you are lazy like me and will most likely be rushed during exams but obviously this wouldn't be acceptable for Algebra class:

Graph each of those functions:
y = (x+9)^(1/3) - (x-9)^(1/3)
y = 3

This method also gives you the visual confirmation of two solutions.

Find the intersections, ±8.94 or 80^(1/2).

Thankfully in my math class all answers to three sig figs are acceptable unless otherwise stated.