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P and Q are the solutions to the quadratic equation x^2+4x+6=0

P^2+Q^2 are the solutions to the quadratic equation x^2+bx+c=0

Find B+C

 Jun 8, 2019
 #1
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+3

Using Vieta's Theorem

 

P + Q  =  -4       (1)

PQ  =  6           (2)

 

Square both sides of  (1)  and we have that

 

P^2 + 2PQ + Q^2 = 16          sub in (2)  for PQ   and we have

 

P^2 + 2(6) + Q^2  = 16

 

P^2 + Q^2  + 12  =  16

 

P^2 + Q^2  =  4

 

So b  = -4

 

And

 

(P^2)(Q^2)  = c

 

(PQ)^2  = c

 

(6)^2  =  c

 

So

 

b = - 4     and c =  36

 

And b + c  =  32

 

 

cool cool cool

 Jun 8, 2019
edited by CPhill  Jun 8, 2019

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