Solve this!!

Any one?
The answer given is 6.

( x - 1 ) ⁿ = x ⁿ - n x ^{n - 1} + ⁿ C ₂ x ^{n - 2} - ⁿ C ₃ x ^{n - 3} + . . . . . . . . ( - 1 ) ⁿ

Clearly , this equation has " n " positive roots .

Comparing co - efficients ,

a ₂ = ⁿ C ₂ ; a ₃ = - ⁿ C ₃

We want , | a ₃ | > | a ₂ |

Or , ⁿ C ₃ > ⁿ C ₂

Or , n > 5

Hence , the least value of " n " is 6 .

Now , a justification -

We see that the product of the roots of the original equation is simply " ( - 1 ) ⁿ " , which indicates that the roots must be of the form ,

" a , 1a " , " b , 1b " , " c , 1c " , 1 , " d , 1d " . . . . . . . . and so on .

But then , the sum of the roots = [ a + 1a ] + [ b + 1b ] + 1 + ....... ≥ 2 + 2 + 1 + ....... > 1 + 1 + 1 ....... > n

Even if there is only one pair of roots " a , 1a " , and the rest are all " 1 " ; then also -

Sum of the roots = [ a + 1a ] + 1 + 1 + 1 ........ ≥ 2 + 1 + 1 + 1 + ......... > n

Hence , all the roots must be " 1 " .

That is why I applied the binomial theorem and compared co -efficients .

Note - ( a + 1a ) ≥ 2 ...........For all " a > 0 " - This is easily verified by AM - GM inequality .

Ricky, your justification isnt right. If the roots are like 6 , 1/2, 1/3, you will still have product of roots=1

A less torturous way to get there is:

Let r_1, r_2,...,r_n be the positive roots.

Now, we are given that r_1+r_2+...+r_n=n

Then, by AM-GM

\frac{r_1+r_2+...+r_n}{n} \ge \sqrt [n] {r_1r_2...r_n} \Rightarrow r_1r_2...r_n \le 1

Equality occurs when r_1 = r_2 =...=r_n=1

As pointed above all the roots are equal .

Now , by AM≥GM

a₂₌\sum{}r_ir_j ≥ (-1)ⁿ C(n,2)

and a₃ = \sum{}r_ir_jr_k≥ (-1)ⁿ C(n,3)

In both cases equality will hold

so we have to find smallest integer n for which

C(n,3) > C(n,2) by inspection n = 6

A doubt

If a polynomial has n equal roots , should we say it has n roots or it has a root of multiplicity n ?

valid doubt that. maybe they should have just said all roots positive.

Thanks to every1!!