# Given that f=x^3+x^2+mx+1 and Sn=x1^n+x2^n+.........,find m if S2=7.

hala718 | Certified Educator

f= x^3 + x^2 + mx +1

Sn = x1^n + x2^n + ....     find m if s2 =7

We know that:

x1+x2+x3= -1/1=-1.....(1)

x1x2+x1x3+x2x3= m......(2)

s2= x1^2 + x2^2 + x3^2 =7

Square(1):

X1^2 + x2^2 + x3^3 + 2(x1x2+x1x3+x2x3)= 1

7+ 2M= 1

2m= -6

m= -3

giorgiana1976 | Student

If Sn=x1^n+x2^n+........., that means that S2 = x1^2+x2^2+x3^2 = 7

Now, we'll substitute the roots into the equation:

x1^3+x1^2+mx1+1 = 0

x2^3+x2^2+mx2+1 = 0

x3^3+x3^2+mx3+1 = 0

If we'll add the 3 equations we'll get:

(x1^3+x2^3+x3^3)+(x1^2+x2^2+x3^2)+m(x1+x2+x3)+3=0

But, from enunciation, x1^2+x2^2+x3^2 = 7.

Also, from Viete's relation, we'll get:

x1^2+x2^2+x3^2 = (x1+x2+x3)^2 - 2(x1*x2+x1*x3+x2*x3)

x1^2+x2^2+x3^2 = (-1)^2 - 2m

7 = 1-2m

7-1 = -2m

6 = -2m

We'll divide by -2:

m = -3

neela | Student

By the relation of roots and coeficients

x1+x2+x3 =-1/1 =-1

x1x2+x2x3+x3x1 =m/1 = m

(x1+x2+x3)^2- (x1^2+x2^2+x3^2)^2  =  2(x1x2+x2x3+x3x1) = 2m.

(-1)^2  - (S2) = 2m

1-7 = 2m.

-6 = 2m . Or

m= -6/2 =-3.

S2 = x1^2+x2^2+s3^2 =