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Old 11-30-2006, 01:38 PM   #1 (permalink)
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Hard algebra question

My girlfriend's math teacher was offering this problem for extra credit:
(7x-2)^(1/3) + (7x+5)^(1/3) = 3
It was due today so it's too late to get extra credit for it, but it's been driving me and my friend completely insane because we are usually pretty proud of our math skills, but this one has us completely stumped. We know (thanks to an online math solver) that the solution is 3/7, but the point was to show steps. We tried using logs but that led us nowhere, and cubing the whole thing leads us to a dead end as well. Can someone at least give me a hint or something? thanks
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Old 11-30-2006, 01:46 PM   #2 (permalink)
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I think what you haveta do is factor out each part, and then get the x's alone, but...I've been sitting here and I haven't the foggiest idea how. lol.

I hope someone else can answer this cause...now it's driving me insane.
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Old 11-30-2006, 04:13 PM   #3 (permalink)
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(7x-2)^(1/3) + (7x+5)^(1/3) = 3


((7x-2)^(1/3))^3 + ((7x+5)^(1/3))^3 = (3)^3
(7x-2) + (7x+5) = 9
14x - 3 = 9
14x = 6
x = 6/14

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Old 11-30-2006, 04:46 PM   #4 (permalink)
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((7x-2)^(1/3))^3 + ((7x+5)^(1/3))^3 = (3)^3

On your left side, you did the power of power while on the right you only did power of the constant..
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Old 11-30-2006, 04:57 PM   #5 (permalink)
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Sorry Dilbert, but it doesn't work like that, because as feelgood mentioned, you only squared the 3, you didn't cube it. You're also basing your solution on the theory that a^n + b^n = (a+b)^n, which is false.
We can plug in some random numbers to prove that this is false:
a=4, b=7, n=3
4^3 + 7^3 = 407
(4 + 7)^3 = 1331
407 is not equal to 1331
Thanks for giving it a shot though
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Old 11-30-2006, 05:22 PM   #6 (permalink)
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Hmmm, well my answer still stands although my methods were definitely wrong.

(7(6/14)-2)^(1/3) + (7(6/14)+5)^(1/3) = 3
(3-2)^(1/3) + (3 + 5)^(1/3) = 3
1^(1/3) + 8^(1/3) = 3
1 + 2 = 3
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Last edited by Dilbert1234567; 11-30-2006 at 05:31 PM.. Reason: spelling
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Old 11-30-2006, 05:45 PM   #7 (permalink)
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Ummm, Dil, 6/14ths reduces to 3/7ths which the OP gave as the answer already...
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Old 11-30-2006, 05:50 PM   #8 (permalink)
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yeah, ok, i'm an idiot... so i'm gona shut up now.
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Old 11-30-2006, 07:19 PM   #9 (permalink)
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complete the cube!
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Old 11-30-2006, 08:56 PM   #10 (permalink)
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(7x-2)^1/3 + (7x+5)^1/3 = 3

=> [(7^1/3)^3)(x^1/3)^3 - (2^1/3)^3)] + [(7^1/3)^3(x^1/3)^3 + (5^1/3)^3) = 3^3

=> [(7)(x) - 2] + [(7)(x) +5] = 27

=> 7x - 2 + 7x + 5 = 27

=> 14x + 3 = 27

=> 14x = 27 - 3

=> 14x = 24

=> x = 24/14

=> x = 12/7


Now let's check the answer:

[7(12/7) - 2]^1/3 + [7(12/7) + 5]^1/3 = 3

=> (12-2)^1/3 + (12+5)^1/3 = 3

=> [(10)^1/3]^3 + [(17)^1/3]^3 = 3^3

=> 10 + 17 = 27

=> 27 = 27

There you go
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Old 11-30-2006, 09:22 PM   #11 (permalink)
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Pornclerk, that doesn't count as checking your answer. If you actually do the math 10^1/3 + 17^1/3 = 4.73 not 3.
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Old 11-30-2006, 09:29 PM   #12 (permalink)
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Quote:
Originally Posted by Zeraph
Pornclerk, that doesn't count as checking your answer. If you actually do the math 10^1/3 + 17^1/3 = 4.73 not 3.
I plugged in the values and did all of the correct math. Tell me what algebra i did wrong and then say it is wrong.
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Old 11-30-2006, 10:34 PM   #13 (permalink)
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Quote:
(7x-2)^1/3 + (7x+5)^1/3 = 3

=> [(7^1/3)^3)(x^1/3)^3 - (2^1/3)^3)] + [(7^1/3)^3(x^1/3)^3 + (5^1/3)^3) = 3^3
I'm not sure you're allowed to break up a power like that. Because like...if you have:

(x-2)^2

=> x^2 -4x + 4

It's kinda hard to relate because of the way we haveta type it out, but I don't see that form there...

Plus, the OP said the answer was 3/7
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Old 11-30-2006, 10:52 PM   #14 (permalink)
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If you plot this function you will see that it only has one zero (solution), which is 3/7.
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Old 12-01-2006, 05:37 AM   #15 (permalink)
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Substitute (7x-2) with a ->

a^1/3 + (a+7)^1/3 = 3

(a+7)^1/3 = 3 - a^1/3

Raise both sides to the power of 3 ->

a+7 = 27 - 27a^1/3 + 9a^2/3 + a

Substitute a^1/3 with b and muck about ->

2b^3 - 9b^2 + 27b - 20 = 0

Which should be solvable. Remember that b = (7x-2)^1/3
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Old 12-01-2006, 10:00 AM   #16 (permalink)
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personally, I like pip's approach. if you want to brute force it, just don't listen to a word i'm saying, because i do math gooder than i spel.
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Last edited by pig; 12-01-2006 at 02:17 PM..
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Old 12-01-2006, 01:38 PM   #17 (permalink)
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Quote:
Originally Posted by pigglet
if you want to brute force it, just rewrite the terms on the left hand side a^(1/3) + b^(1/3) as 1/(a^3) + 1/(b^3).
Pigglet, I love your piggy style, but surely you are not saying that a^1/3 = 1/a^3 ? Because it's a^(-3) = 1/a^3
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Old 12-01-2006, 02:15 PM   #18 (permalink)
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Quote:
Originally Posted by Pip
Substitute (7x-2) with a ->

a^1/3 + (a+7)^1/3 = 3

(a+7)^1/3 = 3 - a^1/3

Raise both sides to the power of 3 ->

a+7 = 27 - 27a^1/3 + 9a^2/3 + a
This is where I don't follow. When I cubed the right side I got the same thing you did, except - a instead of + a, so I get:

a+7 = 27 - 27a^1/3 + 9a^2/3 - a

Then I do b = a^(1/3) so I get:

a + 7 = 27 - 27b + 9b^2 - a

I'll be back, have to go turn something in...
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Old 12-01-2006, 02:19 PM   #19 (permalink)
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Quote:
Originally Posted by Pip
Pigglet, I love your piggy style, but surely you are not saying that a^1/3 = 1/a^3 ? Because it's a^(-3) = 1/a^3
Crap WOMAN, no...ummm....what the dickens are you talking of? seriously, that shit is embarrassing. i can tell its time to wrap this friday off; you should be able to force out the cubic equation without the substitutions, but i suppose doing the math correctly might help...
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Old 12-01-2006, 03:45 PM   #20 (permalink)
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Okay, I'm back, and it looks like a + 7 = 27 - 27b + 9b^2 - a is no good either because now we have two variables
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Old 12-01-2006, 04:58 PM   #21 (permalink)
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Quote:
Originally Posted by Stiltzkin
Okay, I'm back, and it looks like a + 7 = 27 - 27b + 9b^2 - a is no good either because now we have two variables
Stiltzkin,

To step in for my truest math-love pip who spanked my ass earlier, the +a,-a thing was a typo on her part earlier. It's -a, as you found - which gives a coefficient of 2 in front of the b^3 term. For the two variable issue, no prob, substitute a=b^3 and you only have one variable. The problem I'm having with this approach currently is that it will give 3 solutions to the problem, and this problem apparently only has one unique solution. Something is rotten in Denmark, as they say.

Off to cogitate.
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Old 12-01-2006, 09:09 PM   #22 (permalink)
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If you plot 2b^3 - 9b^2 + 27b - 20, it only has one zero, so don't worry about the three possible solutions. I think Pip got it because when I solve 2b^3 - 9b^2 + 27b - 20 = 0 using the calculator (I'll solve it later by completing the cube, my girlfriend says I'm too obsessed with this problem, she's says she's jealous of the problem because I pay more attention to it than to her), you get b = 1, and when you substitute (7x - 2)^1/3 = 1 you get x=3/7, which is the correct solution. So Pip freaking pwns

EDIT: Welp, I looked up several methods for completing the cube:
http://mathworld.wolfram.com/CubicFormula.html
http://mathforum.org/dr.math/faq/faq...quations2.html
http://mathforum.org/dr.math/faq/faq...equations.html

but I don't really follow any of them. I'll have to further investigate this on my own after finals next week
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Last edited by Stiltzkin; 12-02-2006 at 10:30 AM..
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Old 12-02-2006, 11:34 AM   #23 (permalink)
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I feel the love here! Sorry about the typo.

I suck at cubic equations. But since we know the answer is x = 3/7 then we know b = 1 must be an answer to

2b^3 - 9b^2 + 27b - 20 = 0

Which it is, if you sort of look at it. I divided by 2 and broke out (b - 1) just to see if I could still do polynomial divisions and got

(b - 1)(b^2 - 7b/2 +10) = 0

b^2 - 7b/2 +10 = 0 has no real roots. I think guessing answers and breaking out roots was the standard method we used back in highschool. Good times.
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Old 12-02-2006, 06:49 PM   #24 (permalink)
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This is easy... It's only basic high-school algrabra..
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Old 12-02-2006, 08:13 PM   #25 (permalink)
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Completing the cube is basic highschool algebra? Can I see how you would solve it?
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Old 12-03-2006, 02:21 AM   #26 (permalink)
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Wow, i have never taken any algebra courses and had completely ignored this thread until now. I decided to look at it just for fun and to see exactly what was inside and all I can say is holly crap! I have no idea what was going on here but it made my head hurt and my eyes burn. Are you sure this is algebra and not voo-doo ?
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Old 12-03-2006, 02:36 AM   #27 (permalink)
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Hmm. Actually, I just looked at my morning scribbles. I stuffed it up.

If you really have to know, I cubed both sides. Wrong eh. I'll take another shot at it - but yeah, I've egg on my face here.

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Old 12-03-2006, 08:12 AM   #28 (permalink)
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Quote:
Originally Posted by pornclerk
I plugged in the values and did all of the correct math. Tell me what algebra i did wrong and then say it is wrong.
It was breaking up what was in the parentheses and taking the cubed root of each. Here's a simple example to show why it doesn't work:

(2 + 3)^3 = 5^3 = 125

(2^3) + (3^3) = 8 + 27 = 35

If you have any operations within parentheses, you have to do those operations first, then you raise to whatever power, otherwise, everything gest screwed up. Hope that helps.
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Old 12-03-2006, 10:20 AM   #29 (permalink)
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No offense to anyone, but why does everyone think they know math so well when they havn't got a clue?

I mean that's kind of true in general of all subjects, but math seems to have more people who think they know what they're doing when they don't even know how to properly check there work.
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Old 12-03-2006, 11:22 AM   #30 (permalink)
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Quote:
Originally Posted by pornclerk
I plugged in the values and did all of the correct math. Tell me what algebra i did wrong and then say it is wrong.
(7x-2)^1/3 + (7x+5)^1/3 = 3

=> [(7^1/3)^3)(x^1/3)^3 - (2^1/3)^3)] + [(7^1/3)^3(x^1/3)^3 + (5^1/3)^3) = 3^3

Here you assume that (a + b)^n = a^n + b^n which it is not. Unless n = 1. Which it isn't in this case.
You also perform the ^ operation term by term on the left hand side of the equation, which is wrong, but that is probably connected to the previous erroneous assumption.

=> x = 12/7

Now let's check the answer:

[7(12/7) - 2]^1/3 + [7(12/7) + 5]^1/3 = 3

=> (12-2)^1/3 + (12+5)^1/3 = 3


=> [(10)^1/3]^3 + [(17)^1/3]^3 = 3^3

And here you do the ^3 operation term-by-term on the left hand side again, which is wrong. You make the same mistake when you check the answer as when you solve the equation, that's why your answer looks good to you.

Let's check this for real.

(7*12/7-2)^1/3 + (7*12/7+5)^1/3 = 10^1/3 + 17^1/3 = {Punch in on the ol' calculator.} = 2.1544 + 2.57128 > 3 So it's wrong.

Lesson: "Raise to the power of" does not behave like "multiply with".

Last edited by Pip; 12-03-2006 at 11:26 AM..
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Old 12-03-2006, 11:57 AM   #31 (permalink)
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Quote:
Originally Posted by Pip
Lesson: "Raise to the power of" does not behave like "multiply with".
I'm assuming here you mean that "rooting x does not mean raise to the power of x as it's inverse operation"?
Well it does really. The problem comes from that (7x-2)^(1/3) + (7x+5)^(1/3) does not equal [(7x-2)+ (7x+5)]^1/3.

For instance if the problem was [(7x-2)+ (7x+5)]^1/3 = 3 then we could cube both sides and it would work.
Solving it we get 12/7ths like has been shown before, and when we plug it in: [(7(12/7)-2)+ (7(12/7)+5)]^1/3 = 3
cubing it
(12-2)+ (12+5) = 27 and bingo it works.

To recap, (7x-2)^(1/3) + (7x+5)^(1/3) does not equal [(7x-2)+ (7x+5)]^1/3.

In other words when you cube both sides of (7x-2)^(1/3) + (7x+5)^(1/3) = 3 you are not really getting rid of the two ^1/3s you only think you are.
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Old 12-03-2006, 12:29 PM   #32 (permalink)
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I'm assuming here you mean that "rooting x does not mean raise to the power of x as it's inverse operation"?
I think it's more of a non distributive thing, i.e. (a+b)^c != a^c + b^c
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Old 12-03-2006, 12:31 PM   #33 (permalink)
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No, I actually meant how it operates on functions and such. Like when you multiply, say, a polynome with, say, the number 3, then you simply multiply each term in the polynome with 3. But when you raise a polynome with the power of 3 ("cube it") then you do not simply raise each term in the polynome with the power of 3. You multiply the polynome with itself twice instead. But you knew that already. I was just unclear.

EDIT: Filtherton got it, and explained better than me.
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Old 12-07-2006, 04:01 AM   #34 (permalink)
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I guess I used to know math - well enough to scrape through a math/phys degree. But I've done so much since then that it's all vanished. Occasionally I pull out a bit of astrophysics just to see how weird it looks.

Oh - and theres an "i" in "their". Yes? : >
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Old 12-07-2006, 04:38 AM   #35 (permalink)
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yeah, i can't really say anything on this one...i'd say i've got a fairly decent amount of math behind me, but holy shit on a pogo stick did i fuck up some simple shit here. i thought about writing back some smartass retort to old zeraph, but i mean - one does not build a house on a foundation of sand. the only thing that strikes me funny is the old "I don't mean to offend anyone, but..." line. Don't get me wrong, I use it all the time myself, but its pretty much a qualifier that what I'm about to say is offensive, and I know it. alack and alas, good times.
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Old 12-07-2006, 09:10 PM   #36 (permalink)
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Quote:
Originally Posted by pigglet
yeah, i can't really say anything on this one...i'd say i've got a fairly decent amount of math behind me, but holy shit on a pogo stick did i fuck up some simple shit here. i thought about writing back some smartass retort to old zeraph, but i mean - one does not build a house on a foundation of sand. the only thing that strikes me funny is the old "I don't mean to offend anyone, but..." line. Don't get me wrong, I use it all the time myself, but its pretty much a qualifier that what I'm about to say is offensive, and I know it. alack and alas, good times.
haha I know
Mostly I used that line though to make a point that I wasn't attacking anyone specifically.
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