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phamthephong100

Member

Registered: 2007-10-13

Posts: 8

Who can solve this problem ? I can . How about you????? :)

(x+2)*(x+2)+(x+3)*(x+3)*(x+3)+(x+4)*(x+4)*(x+4)*(x+4) = 2
x=?????????
show me !!!!!!!

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: Who can solve this problem ? I can . How about you????? :)

x = -3

(x+2) * (X+2) = 1
(x+3)(x+3)(X+3) = 0
(X+4)(X+4)(X+4)(X+4) = 1

1+0+1 = 2

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Who can solve this problem ? I can . How about you????? :)

If you let

, the equation becomes less messy:

Thus you can easily see that

(i.e.

) is one solution. To find the other solutions, expand and solve for y, then substitute back to find x.

The full solution set is

Last edited by JaneFairfax (2008-01-18 05:10:06)

phamthephong100

Member

Registered: 2007-10-13

Posts: 8

Re: Who can solve this problem ? I can . How about you????? :)

your solution is quite stupid
why dont you let a=x+4 that 's much easier :lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol::lol:

Last edited by phamthephong100 (2008-01-26 17:30:48)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Who can solve this problem ? I can . How about you????? :)

If you use

you’ll get a quartic equation with nonzero constant term, whereas my quartic equation above has a zero constant term. Which is easier to solve?

To illustrate what I mean, if you use your method

, you get

If you use my method

, you get

Now which of the two equations is harder to solve, and hence whose method is more stupid? I am going to seek an independent opinion for this.

Last edited by JaneFairfax (2008-01-27 01:22:22)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Who can solve this problem ? I can . How about you????? :)

jane your equation is much easier to solve as a result of the non zero constant

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The Beginning Of All Things To End.
The End Of All Things To Come.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Who can solve this problem ? I can . How about you????? :)

jane your equation is much easier to solve as a result of the non zero constant

You mean as a result of zero constant.

Well, I knew someone would agree with me. Thank you, Luca! Grazie, grazie!

Last edited by JaneFairfax (2008-01-27 01:39:14)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Who can solve this problem ? I can . How about you????? :)

>.> yeh that's what i meant.

prego ^^

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The Beginning Of All Things To End.
The End Of All Things To Come.

phamthephong100

Member

Registered: 2007-10-13

Posts: 8

Re: Who can solve this problem ? I can . How about you????? :)

if you use y=x+3 ,you have to
(y+1)*(y+1)*(y+1)*(y+1)= y*y*y*y+4*y*y*y+6*y*y+4*y+1(easier ???)
my method
       y*y*y*y+(y-1)*(y-1)*(y-1)+(y-2)*(y-2)-2=0
<=>y*y*y*y+ y*y*y-2*y*y-y+1=0
<=>y*y*y*(y+1)-y*(y+1)-(y+1)(y-1)=0
<=>y*y*(y*y+y-1)-(y*y+y-1)=0
<=>y*y-1=0 or y*y+y-1=0

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Who can solve this problem ? I can . How about you????? :)

I have no idea what the ph*ck PhatThePhong100 is talking about but take a look at this:

http://www.mathhelpforum.com/math-help/ … solve.html

It would appear that PhatIhePhong is “ignorant” and “immature … because he has a solution that he thinks it is the best”. :lol::lol::lol::lol::lol::lol::lol:

phamthephong100

Member

Registered: 2007-10-13

Posts: 8

Re: Who can solve this problem ? I can . How about you????? :)

Step by step

*1st method:(JaneFairfax)
Let

we have

             

             

             

It is too long to develop

and if you do so you may make some mistakes while you developing .
*2nd method
Let

We have

With this method you dont have to develop any

to something
I think that the point that make the 2nd method easier.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Who can solve this problem ? I can . How about you????? :)

It is too long to develop

.

Oh yeah? You don’t know the binomial theorem, then, it seems.

What is so difficult about that? With x = 1, the form is even simpler:

YOU are clearly the one who is being ignorant here. Go and learn some useful mathematics before you start throwing your weight around.