62
Lokesh Verma
·2009-01-08 11:13:19
t=3x-3
t/x = 3-3/x
x1/3 + (t-x)1/3 = (4t)1/3
1+ (t/x-1)1/3 = (4t/x)1/3
taking cube
1+ 3(t/x-1)1/3 + 3(t/x-1)2/3 + (t/x-1) = 4t/x
3(t/x-1)1/3 + 3(t/x-1)2/3 = 3t/x
(t/x-1)1/3 + (t/x-1)2/3 = t/x
(t/x-1) + (t/x-1)2 +3(t/x-1)(t/x) = (t/x)3
(t/x-1) + (t/x-1)2 +3(t/x-1)(t/x) = (t/x)3
(t/x-1)t/x +3(t/x-1)(t/x) = (t/x)3
4(t/x-1)(t/x) = (t/x)3
Thus , t/x=0 or
4(t/x-1) = (t/x)2
(t/x)2 -4t/x +4 =0
t/x=2
thus solved :)
62
Lokesh Verma
·2009-01-08 11:26:33
t/x = 3-3/x
3-3/x= 0
x=1
3-3/x =2
3/x=1
x=3
so x=1, 3 are the solutions!
24
eureka123
·2009-01-09 02:58:22
Thanks for reply..
Just one question.:: Can these type of questions come in JEE ?? I mean to say that should I practice more of such type or not. ??
62
Lokesh Verma
·2009-01-09 03:04:57
frankly i think i have messed up the solution.. so it is looking dirty and tough..
JEee will have more simpler problems (or which are solved in lesser no of steps.!)
62
Lokesh Verma
·2009-01-09 03:10:11
x1/3 + (2x-3)1/3 = (12x-12)1/3
take cube
x+2x-3 + {x(2x-3)}1/3{x1/3 + (2x-3)1/3} = 12x-12
thus,
x+2x-3 + {x(2x-3)}1/3 (12x-12)1/3= 12x-12
thus,
{x(2x-3)}1/3 (12x-12)1/3 = 9x-9
Now this is simple? (oops simpler?)
To try to make the solution simple, i made it more difficult :D
341
Hari Shankar
·2009-01-09 05:18:32
Let a = x1/3 and b = (2x-3)1/3
Then the equation is
a1/3+b1/3 - 41/3(a+b)1/3 = 0
We know that if p+q+r =0, then p3+q3+r3 = 3pqr
Hence a+b-4(a+b) = -3.(4ab)1/3(a+b)1/3 or
(a+b) = (4ab)1/3(a+b)1/3
Now either a+b = 0 i.e. x = 1 or
(a+b)2/3 = (4ab)1/3
Cubing both sides, we get
(a+b)2 = 4ab or (a-b)2 = 0 or a = b
Hence x = 2x-3 giving x = 3 as the other solution
62
Lokesh Verma
·2009-01-09 05:20:53
great man ....
I was trying to simplify it.. and made a big mess of it :D
Godo that you simplified it :)
341
Hari Shankar
·2009-01-09 05:23:35
There are several of these kind of problems. One that comes to mind immediately is:
Solve (6-x)4+(x-8)4 = 16
Unless you spot that this method can be used, it will be a very messy problem.
I will see if I can hunt down other problems of this variety
341
Hari Shankar
·2009-01-09 05:27:00
and since this is eureka's thread I want to just say that see we will be eager to help you out bcos u r a bro (applies to sis also) who is on the same path. But when we put effort and then no one says a thing it will be like wasted. Now Nishant brother said something so it makes me look if i can add something more. You get more cooperation just by being nice
341
Hari Shankar
·2009-01-09 06:10:30
Ok a similar problem:
Solve (x2+x-2)3+(2x2-x-1)3 = 27(x2-1)3