prove it if possible?

Question

prove that the polynomial
(a-b)^(n)+nk(a-b)=1
has no real solutions if n>2.(Here a , b & k are real numbers.)

Answer 1

This is manifestly false if n is odd.
Let's let x = a-b. Then x^n +nkx-1= 0.
If n is an odd integer this polynomial has at least
1 real zero by the intermediate value theorem.
In fact, it's false for any value of n.
Take k = 1 and look at p(x) = x^n + nx -1,
Note that p(0) = -1 < 0 but p(1) = n. Since n > 2,
this is positive. So there is a real root of p(x)
between 0 and 1, again by the intermediate
value theorem.

Answer 2

its not true that for n>2, (a-b)^(n)+nk(a-b)=1has no real solutions.

set, a=5, b=3, n=3 and k=-7/6
u get a valid answer.

also u can prove it like this way:
n=m
=>(a-b)^(m)+mk(a-b)=1
and
n=m+1
=>(a-b)^(m+1)+(m+1)k(a-b)=1

thus,
(a-b)^(m+1)-(a-b)^m+k(a-b)=0
=>x*x^m-x^m+kx=0 [a-b=x]
=>(1-mkx)(x-1)+kx=0
=>x-mkx^2-1+mkx+kx=0
=>mkx^2-(mk+k+1)x+1=0

for this eqn,
D=(mk+k+1)^2-4mk

certain conditions like m>2, D will not be negative.
thus, for n>2 u can get certain real solutions(as showed earlier)

Answer 3

First if we say that n is 1 then the equation would look some thing like that

(a-b)+(ka-kb)=1 and then we say that k is 0 so it would look like that

a-b=1

so a=b+1

Answer 4

(a-b)^(n)+nk(a-b)=1
(a-b)[ (a-b)^{n-1} +nk ] =1
what are the variables? a, b, or k
or allof them?

Answer 5

Rajesh,

I really don't give a darn!!

Answer 6

ur mistaken dude...it is always not necessary...

Answer 7

(a-b)^(n) + nka - nkb = 1
!!!!!!!!!!!!!!!!