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Let Q(x) = x^2+bx+c where b and c are integers.

If Q\left((1+\sqrt 2)^8\right)=0 determine b+c .

waffles  Jan 4, 2018
 #1
avatar+92857 
+1

(1  +   √2)^8  evaluates to  577 + 408√2

 

So.....this is a root....and so is the conjugate, 577 - 408√2

 

And the sum of the roots  = -b

 

So

 

577  +  577  =  - b  ⇒  1154  ⇒  b  =  -1154

 

And the product of the roots  = c

 

So

 

(  577 + 408√2)  (  577 - 408√2)   =  c   =   ( 332929  - 332928)  =   1

 

So

 

b  +  c  =   -1154  +   1   =   -1153

 

 

cool cool cool

CPhill  Jan 4, 2018

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