If p , q & r are the roots of polynomial x³+4x²-3x+5 =0 Find the value of (p⁵+q⁵+r⁵) + (p⁴+q⁴+r⁴) – (p²+q²+r²)²
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.
Sir please tell me that is there any formula to calculate P n
means how can we calculate P 8 or P 9 With respect To S n
Log in to reply
Mehul, for a k t h degree polynomial with roots a 1 , a 2 , a 3 . . . a k , there will be S 1 , S 2 , S 3 . . . S k . This question is for 3 roots therefore there are S 1 , S 2 , S 3 . Then, P n is given by:
P n = { S 1 P n − 1 − S 2 P n − 2 + S 3 P n − 3 . . . + ( − 1 ) n − 1 n S n S 1 P n − 1 − S 2 P n − 2 + S 3 P n − 3 . . . + ( − 1 ) k − 1 S k P n − k n ≤ k n > k
For this question:
P 6 = − 4 ( P 5 ) + 3 ( P 4 ) − 5 ( P 3 ) = − 4 ( − 5 6 9 ) + 3 ( 1 6 2 ) − 5 ( − 4 2 ) = 2 0 0 5
P 7 = − 4 ( 2 0 0 5 ) + 3 ( − 5 6 9 ) − 5 ( − 1 6 2 ) = − 7 1 2 3
P 8 = − 4 ( − 7 1 2 3 ) + 3 ( 2 0 0 5 ) − 5 ( − 5 6 9 ) = 2 5 3 2 2
P 9 = − 4 ( 2 5 3 2 2 ) + 3 ( − 7 1 2 3 ) − 5 ( 2 0 0 5 ) = − 8 9 9 4 4
Log in to reply
A big thanks to you sir!!
Log in to reply
@Mehul Chaturvedi – You don't even need to remember the formula. Just plug in the roots (in the polynomial) in succession and add, you'll get the same result.
You can also multiply the polynomial by a power of x , since this won't change anything (Although you are introducing more roots, but that doesn't matter) and then plug in the roots and then add to get the sum.
Log in to reply
@Aneesh Kundu – Would you please clarify more?
Log in to reply
@Mehul Chaturvedi – For simplicity lets just assume that there are 3 roots(although this can be generalised easily). Let S n denote the sum of the n th powers of the roots, i.e. S n = α n + β n + γ n let α , β , γ be the roots of f ( x ) = a x 3 + b 2 + c x + d = 0 Since α , β , γ are the roots of f ( x ) = 0 , f ( α ) = a α 3 + b α 2 + c α + d = 0 f ( β ) = a β 3 + b β 2 + c β + d = 0 f ( γ ) = a γ 3 + b γ 2 + c γ + d = 0 Adding all the eqns, we get a S 3 + b S 2 + c S 1 + 3 d = 0 What I did above can also be stated as plugging the roots in succession and then adding.
Well, if we know S 2 and S 1 we can easily calculate S 3 . But we could calulate S 3 even without that, so how is that useful? In the next step we draw some useful results and try to generalize a bit. We started with f ( x ) = a x 3 + b 2 + c x + d = 0 what if multiply both sides by x n − 3 ? Lets just try that a x n + b n − 1 + c x n − 2 + d x n − 3 = 0 Since we just multiplied the equation by a number, this still hasn't affected anything(although we have introduced more roots), this means that α , β , γ are still its roots.
Now we plug-in the roots in succession and add.
a S n + b S n − 1 + c S n − 2 + d S n − 3 = 0
This is a very useful result and it can help us in calculating S n if we know S n − 1 , S n − 2 , S n − 3 (and also the coefficients of the polynomial).
Hope this helped.
Log in to reply
@Aneesh Kundu – Thanks for your explanation. I will try it out in the next question.
From which book,You read this?
Log in to reply
@Mehul Chaturvedi – I googled and did some thinking. http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums. Read also Vieta's formulas.
I did same
Problem Loading...
Note Loading...
Set Loading...
This problem can be solved using Newton's Sums method.
Let S 1 = p + q + r = − 4 , S 2 = p q + q r + r p = − 3 , S 3 = p q r = − 5 and P n = p n + q n + r n , where n = 1 , 2 , 3 .... Then P n are given by:
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ P 1 = S 1 P 2 = S 1 P 1 − 2 S 2 P 3 = S 1 P 2 − S 2 P 1 + 3 S 3 P 4 = S 1 P 3 − S 2 P 2 + S 3 P 1 P 5 = S 1 P 4 − S 2 P 3 + S 3 P 2 = − 4 ( − 4 ) − 2 ( − 3 ) = − 4 ( 2 2 ) + 3 ( − 4 ) + 3 ( − 5 ) = − 4 ( − 1 5 5 ) + 3 ( 2 2 ) − 5 ( − 4 ) = − 4 ( 5 4 6 ) + 3 ( − 1 5 5 ) − 5 ( − 2 2 ) = − 4 = 2 2 = − 1 5 5 = 5 4 6 = − 2 6 3 9
Therefore,
( p 5 + q 5 + r 5 ) + ( p 4 + q 4 + r 4 ) − ( p 2 + q 2 + r 2 ) 2
= P 5 + P 4 − P 2 2 = − 2 6 3 9 + 5 4 6 − ( 2 2 2 ) = − 2 5 7 7