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Let f(x)=(x^2+6x+9)^{50}-4x+3, and r_1,r_2,...r_100 be the roots of f(x).

Compute (r_1+3)^100+(r_2+3)^100+...+(r_100+3)^100

 Apr 16, 2022

Best Answer 

 #1
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Note that 

f(x)=(x2+6x+9)504x+3=((x+3)2)504x+3=(x+3)1004x+3

 

For each of i=1,2,,100, since x=ri is a root of f(x):

f(ri)=0(ri+3)100=4ri3

 

Now, using the equation above,

(r1+3)100+(r2+3)100++(r100+3)100=(4r13)+(4r23)++(4r1003)=4(r1+r2++r100)300

 

By Vieta's formula, we have:

r1+r2++r100=coefficient of x99 in f(x)coefficient of x100 in f(x)

 

Using the binomial theorem, we can expand (x+3)100.

(x+3)1004x+3=x100+300x99+

 

Therefore, 

r1+r2++r100=3001=300

 

Recall that (r1+3)100+(r2+3)100++(r100+3)100=4(r1+r2++r100)300. Then:

(r1+3)100+(r2+3)100++(r100+3)100=4(300)300=1500

 Apr 16, 2022
 #1
avatar+9675 
+2
Best Answer

Note that 

f(x)=(x2+6x+9)504x+3=((x+3)2)504x+3=(x+3)1004x+3

 

For each of i=1,2,,100, since x=ri is a root of f(x):

f(ri)=0(ri+3)100=4ri3

 

Now, using the equation above,

(r1+3)100+(r2+3)100++(r100+3)100=(4r13)+(4r23)++(4r1003)=4(r1+r2++r100)300

 

By Vieta's formula, we have:

r1+r2++r100=coefficient of x99 in f(x)coefficient of x100 in f(x)

 

Using the binomial theorem, we can expand (x+3)100.

(x+3)1004x+3=x100+300x99+

 

Therefore, 

r1+r2++r100=3001=300

 

Recall that (r1+3)100+(r2+3)100++(r100+3)100=4(r1+r2++r100)300. Then:

(r1+3)100+(r2+3)100++(r100+3)100=4(300)300=1500

MaxWong Apr 16, 2022
 #2
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 Good Job!

Guest Apr 21, 2022

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