11302006, 01:38 PM  #1 (permalink) 
Junkie

Hard algebra question
My girlfriend's math teacher was offering this problem for extra credit:
(7x2)^(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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The most important thing in this world is love. 
11302006, 04:13 PM  #3 (permalink) 
Devils Cabana Boy
Location: Central Coast CA

(7x2)^(1/3) + (7x+5)^(1/3) = 3
((7x2)^(1/3))^3 + ((7x+5)^(1/3))^3 = (3)^3 (7x2) + (7x+5) = 9 14x  3 = 9 14x = 6 x = 6/14
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11302006, 04:46 PM  #4 (permalink) 
Free Mars!
Location: I dunno, there's white people around me saying "eh" all the time

((7x2)^(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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11302006, 04:57 PM  #5 (permalink) 
Junkie

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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The most important thing in this world is love. 
11302006, 05:22 PM  #6 (permalink) 
Devils Cabana Boy
Location: Central Coast CA

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 (32)^(1/3) + (3 + 5)^(1/3) = 3 1^(1/3) + 8^(1/3) = 3 1 + 2 = 3
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Donate Blood! "Love is not finding the perfect person, but learning to see an imperfect person perfectly." Sam Keen Last edited by Dilbert1234567; 11302006 at 05:31 PM.. Reason: spelling 
11302006, 08:56 PM  #10 (permalink) 
Insane
Location: Ontario, Canada

(7x2)^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 => (122)^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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11302006, 09:29 PM  #12 (permalink)  
Insane
Location: Ontario, Canada

Quote:
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Who wants a twig when you can have the whole tree? 

11302006, 10:34 PM  #13 (permalink)  
Tilted
Location: South Carolina

Quote:
(x2)^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 

12012006, 05:37 AM  #15 (permalink) 
Likes Hats
Location: Stockholm, Sweden

Substitute (7x2) 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 = (7x2)^1/3 
12012006, 10:00 AM  #16 (permalink) 
pigglet pigglet
Location: Locash

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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You don't love me, you just love my piggy style Last edited by pig; 12012006 at 02:17 PM.. 
12012006, 01:38 PM  #17 (permalink)  
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Location: Stockholm, Sweden

Quote:


12012006, 02:15 PM  #18 (permalink)  
Junkie

Quote:
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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The most important thing in this world is love. 

12012006, 02:19 PM  #19 (permalink)  
pigglet pigglet
Location: Locash

Quote:
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12012006, 04:58 PM  #21 (permalink)  
pigglet pigglet
Location: Locash

Quote:
To step in for my truest mathlove 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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You don't love me, you just love my piggy style 

12012006, 09:09 PM  #22 (permalink) 
Junkie

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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The most important thing in this world is love. Last edited by Stiltzkin; 12022006 at 10:30 AM.. 
12022006, 11:34 AM  #23 (permalink) 
Likes Hats
Location: Stockholm, Sweden

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. 
12032006, 02:21 AM  #26 (permalink) 
Crazy
Location: The Darkest Parts Of Places Unknown

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 voodoo ?

12032006, 08:12 AM  #28 (permalink)  
Junkie

Quote:
(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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12032006, 10:20 AM  #29 (permalink) 
Banned
Location: The Cosmos

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. 
12032006, 11:22 AM  #30 (permalink)  
Likes Hats
Location: Stockholm, Sweden

Quote:
=> [(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 => (122)^1/3 + (12+5)^1/3 = 3 => [(10)^1/3]^3 + [(17)^1/3]^3 = 3^3 And here you do the ^3 operation termbyterm 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*12/72)^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; 12032006 at 11:26 AM.. 

12032006, 11:57 AM  #31 (permalink)  
Banned
Location: The Cosmos

Quote:
Well it does really. The problem comes from that (7x2)^(1/3) + (7x+5)^(1/3) does not equal [(7x2)+ (7x+5)]^1/3. For instance if the problem was [(7x2)+ (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 (122)+ (12+5) = 27 and bingo it works. To recap, (7x2)^(1/3) + (7x+5)^(1/3) does not equal [(7x2)+ (7x+5)]^1/3. In other words when you cube both sides of (7x2)^(1/3) + (7x+5)^(1/3) = 3 you are not really getting rid of the two ^1/3s you only think you are. 

12032006, 12:31 PM  #33 (permalink) 
Likes Hats
Location: Stockholm, Sweden

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. 
12072006, 04:01 AM  #34 (permalink) 
Junkie
Location: Melbourne, Australia

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? : > 
12072006, 04:38 AM  #35 (permalink) 
pigglet pigglet
Location: Locash

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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12072006, 09:10 PM  #36 (permalink)  
Banned
Location: The Cosmos

Quote:
Mostly I used that line though to make a point that I wasn't attacking anyone specifically. 

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