JEE Advanced algebra problem
Let there be a polynomial a x 4 + b x 3 + c x 2 + d x + e such that a + e = 5 , a e = 6 , e > a . And a , b , c , d and e are all real numbers. Let α , β , γ and δ be roots of the above polynomial. Find
( b 2 − a c ) S 1 2 S 1 0 + ( b c − a d ) S 1 2 S 9 + ( b d − a e ) S 1 2 S 8 + b e S 1 2 S 7 ,
where S n = α n + β n + γ n + δ n , where n is a positive integer.
The answer is 4.
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1 solution
exactly and short solution
How you resolved the whole polynomial into a recursive equation?
Using polynomial equation we can establish a recurrence relation:
a S n + b S n − 1 + c S n − 2 + d S n − 3 + e S n − 4 = 0 And required expression can be rearranged as: S 1 2 b ( b S 1 0 + c S 9 + d S 8 + e S 7 ) − S 1 2 a ( c S 1 0 + d S 9 + e S 8 )
= S 1 2 b ( − a S 1 1 ) − S 1 2 a ( − a S 1 2 − b S 1 1 )
= a 2 = 4
Using the condition a + e = 5 and a e = 6 and a < e we can find a = 2 .