A Polynomial Problem with Vieta's Rules
r 2 r 3 r 4 r 5 + r 1 r 3 r 4 r 5 + r 1 r 2 r 4 r 5 + r 1 r 2 r 3 r 5 + r 1 r 2 r 3 r 4
Let the roots of P ( x ) = 4 x 5 − 1 3 x 3 − 1 2 x + 7 be r 1 , r 2 , r 3 , r 4 and r 5 , then find the value of the expression above.
The answer is -3.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Why complicating it so much? :P
Log in to reply
I don't see any complications :-P
You can directly use Viete's Formula here. Inspection of the given expression reveals that it is sum of the roots of given quintic polynomial taken 4 at a time. This can be easily verified as every term in the given sum is a product of distinct roots and number of terms is 5 which is equal to ( 4 5 ) .
Log in to reply
I knew that but I posted this solution just for the sake of variety.
This is simply Vieta's formula whose product of the sum of their roots taken 4 at a time. We know that if a , b , c , d , e are the roots of f x 5 + g x 4 + h x 3 + i x 2 + j x + k , the value of a b c d + b c d e + c d e a + d e a b + e a b c = j / a
r 2 r 3 r 4 r 5 + r 1 r 3 r 4 r 5 + r 1 r 2 r 4 r 5 + r 1 r 2 r 3 r 5 + r 1 r 2 r 3 r 4 ( r 1 1 + r 2 1 + r 3 1 + r 4 1 + r 5 1 ) r 1 r 2 r 3 r 4 r 5 P ( x ) = 4 x 5 − 1 3 x 3 − 1 2 x + 7 r 1 r 2 r 3 r 4 r 5 = − 4 7 x = y 1 ⇒ P ( x ) = 7 y 5 − 1 2 y 4 + − 1 3 y 2 + 4 ∑ y = ( r 1 1 + r 2 1 + r 3 1 + r 4 1 + r 5 1 ) = 7 1 2 ( r 1 1 + r 2 1 + r 3 1 + r 4 1 + r 5 1 ) r 1 r 2 r 3 r 4 r 5 = ( 7 1 2 ) ( − 4 7 ) ⇒ − 3