Question

Given f(x) = x^3+9x^2-5x-33 , find the sum of the squares of its zeros

Answer 1

graph it, find the x intercepts, then you can take the sum of its squares

Answer 2

what does it mean sum of squares

Answer 3

@perl

Answer 4

@PaxPolaris

Answer 5

first find the x intercepts

Answer 6

you can use wolfram

Answer 7

if the zeroes are a, b and c

find\(a^2+b^2+ c^2\)

Answer 8

http://www.wolframalpha.com/input/?i=roots+of+f%28x%29+%3D+x^3%2B9x^2-5x-33

Answer 9

we were only allowed to use the sums and product

Answer 10

sum is -9 and product is -33

Answer 11

See
http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums

Answer 12

use Vieta's formulas and Newton sums

Answer 13

\[(a+b+c)^2 \\ = \left[ \left( a+b \right) +c\right]^2\\ =\left( a^2+b^2+2ab \right)+c^2+2c(a+b)\\=a^2+b^2+c^2+2(ab+ac+bc)\]

Answer 14

Nice :)

Answer 15

\[a^2 +b^2+c^2=(a+b+c)^2-2(ab+ac+bc)\] .... not sure if this helps though?

Answer 16

Yes it does @PaxPolaris thank you!

Answer 17

how do you know (ab+ac+bc) @Shruti_ ?

Answer 18

when you foil and factor

Answer 19

what would you substitute in for a b and c

Answer 20

a+b+c = -9

Answer 21

right but what about the second half

Answer 22

\[\large f(x) = c_3x^3 + c_2x^2 + c_1x^1 + c_0\]

\(\large a+b+c = -\dfrac{c_2}{c_3}\)

\(\large ab+bc+ca = \dfrac{c_1}{c_3}\)

\(\large abc =- \dfrac{c_0}{c_3}\)

Answer 23

how about -33/c + -33/b + -33/a

Answer 24

thats what I got! where would i go from there?

Answer 25

dunno :s

Answer 26

Thanks for your help! @PaxPolaris

Answer 27

\[a^2 +b^2+c^2=(a+b+c)^2-2(ab+ac+bc) = (-9)^2 -2(-5)^2\]

simplify and you're done right ?

Answer 28

How would it be -5? @rsadhvika

Answer 29

VIeta's formula

Answer 30

ohhhh okay thank you for elaborating!! @rsadhvika

Answer 31

@rsadhvika how did you get that ?

Answer 32

deriving them is easy, just compare the coefficients of the given polynomial and the factored form

Answer 33

say a,b,c are the zeroes of given polynomial :
\[\large x^3+9x^2-5x-33 = (x-a)(x-b)(x-c)\]

Answer 34

compare the coefficients of x^2 term both sides you will get :
\(\large 9 = -a-b-c\)

Answer 35

compare the coefficients of x term both sides you get :
\(\large -5 = ab+bc+ca\)

Answer 36

similarly comparing constant terms you get
\(\large -33 = -abc\)

Answer 37

you can actually solve for a,b,c from those three equations

Answer 38

Indeed.

Answer 39

ok but someone above said to use a different formula

Answer 40

but a,b and c look crazy ... the sum of squares is a nice integer @perl

Answer 41

true, thats the reason we use newton sums

Answer 42

newwton sums is very useful for finding sums of higher powers

Answer 43

so 91 your answer @Shruti_

Answer 44

Yes!