If equation xn - nxn-1 + a2xn-2 + a3xn-3 + ......an-1x + (-1)n = 0 has n positive roots, then least value of n for which a2 + a3 is negative, is ?
1( 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 .
341Ricky, 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
21As pointed above all the roots are equal .
Now , by AM≥GM
a2=\sum{}rirj ≥ (-1)n C(n,2)
and a3 = \sum{}rirjrk≥ (-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
21A doubt
If a polynomial has n equal roots , should we say it has n roots or it has a root of multiplicity n ?
341valid doubt that. maybe they should have just said all roots positive.
