# Can you solve it?

Discussion in 'General Off-Topic Chat' started by MaHe, Apr 21, 2008.

# Can you solve it?

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1. ### MaHeone lazy schmo

Member
2
Aug 4, 2006
Maribor
Here's a little math problem. Let's see if you can actually solve it.
While I know the solution, I have no idea how to reach it by calculation (I guessed ).

a = c2d + c2
b = cd + a:b
c = a - b - cd
d = ((c+c+c):c):c

Figure out the values of a, b, c and d. Only one solution is possible.
And just to clarify ":" means to divide.

EDIT: Holy crap, wrong forum. >_

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Aug 29, 2007

3. ### xcaliburGbatemp's Chocolate Bear

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Jun 2, 2007
Sacred Heart
Then why the hell did you bother posting?

Anyways, I would imagine you would have to put answers in terms of other words.
e.g c+c+c=3c, 3c:c = 3, 3:c=d

and just workd your way up by just putting stuff in terms of others.

4. ### MaHeone lazy schmo

Member
2
Aug 4, 2006
Maribor
This is not my homework, I made the equations ... Just modified one that I was able to solve and added a twist ...
Warning: Spoilers inside!
And please, can a mod move this to another forum?

5. ### SonicslasherIn Law we trust.

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3
Dec 5, 2005
Richmond, VA
muV 2 tIsTiNG aReA!!!111

6. ### PsyfiraCredit: 0ml. Insert tea to continue

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3
Dec 31, 2003
England
Been a while, heh,I just proved that b=b Useful much?

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3
May 14, 2006
For the calculation, just replace the variables with equivalents. since d = *long equation here* just plug it into a so you have two variables etc im too lazy

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1
Aug 31, 2007
England

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2
Nov 4, 2006
England land
Yes.

10. ### deathfisaroNarcistic Deathfisaro Fan

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2
Mar 16, 2007
Vancouver, BC
a = c^2d + c^2
b = cd + a/b
c = a - b - cd
d = ((c+c+c)/c)/c

d = 3/c
substitute d = 3/c
a = 3c + c^2 (1)
b = 3+ a/b -> b^2 = 3b + a, a = b^2 - 3b (2)
c = a - b - 3 (3)

combine (1) and (2)
3c + c^2 = b^2 - 3b (4)

substitute (2) into (3)
c = b^2 - 3b - b - 3 = b^2 - 4b - 3 (5)

substitute (5) into (4)
(b^2 - 4b - 3)*(b^2 - 4b) = b^2 - 3b

You get
b(b^3 - 8b^2 + 12b + 15) = 0,
so if b = 0, then c^2d + c^2 = c^2d, so c =0, then d = 0/0 which isn't desirable so b=/=0
That means b^3 - 8b^2 + 12b + 15 = 0
because 15 has factors of 1,3,5,15, try factoring with (b-1), doesn't work. (b-3) doesn't work,
Factor with (b-5) you get (b-5)(b^2-3b-3) = 0
so if b = 5, then c = 25-20-3 = 2 from (5), then a = 3*2 + 2^2 = 10 from (1), and d = 3/2.
(The only way to factor out b^2 - 3b - 3 is factoring by (b-1) and (b-3), and we've already done that 3 lines above and failed)

The reason you factor with (b-x) where x is a factor of 15 is because the only way you get 15 at the end is by multiplying (b-x) by some other constant
(For example, from (b^2 - 3b - 3), 15 has to be -x * -3 otherwise you get a multiple of "b", not a constant)

Before university, there may be one acceptable solution but I see a=0, b=0, c=0, d=0/0 as an answer too.

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Feb 23, 2006

12. ### SonicslasherIn Law we trust.

Member
3
Dec 5, 2005
Richmond, VA
Divide by zero

13. ### deathfisaroNarcistic Deathfisaro Fan

Member
2
Mar 16, 2007
Vancouver, BC
Code got busted there, should have casted to double or float, because mathematically 1 != ((3 * 2) / 2) / 2.

edit: tried bolding the code, didn't work =P

14. ### AzimuthChicken Teriyaki Boy!

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1
Feb 23, 2006
yeah, my bad, what a stupid mistake. This is what working in scripting languages does to a person, you start taking these small things for granted

15. ### Dylan100 MILES AND RUNNIN'

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Mar 7, 2008
Sydney, Australia

16. ### BORTZTired of being the good guy

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Dec 2, 2007
Pittsburgh

i tried that before remember?

17. ### WestsideSogdiana

Member
3
Dec 18, 2004
Guantanamo bay
I also would like to contribute to this wonderful thread:
- Find the volume of the solid generated by revolving the plane region bounded by the equations :
y=(1+x^2)^-(1/2) ; y=0; x=1; x=-1
Revolved around X-axis, using the disc method.

18. ### HarpuiaGBAtemp Fan

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Jun 6, 2007
My House
I used a calculator and got about 4.935

19. ### gizmo_galQWEEN of the RadioActive Force!!!

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Apr 10, 2008
1st dimension, Earth
.... Um, ...uh...

Sorry, I was never any good at algebra...

20. ### xcaliburGbatemp's Chocolate Bear

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Jun 2, 2007
Sacred Heart
Do you square it before or after you integrate it?
I always forget when you square y...