In the polynomial f(x) =2*(x)^4 - 49*(x)^2 +54, what is the product and sum of the roots

• SUM OF ROOTS = (-1)^1 COEFFICIENT OF X^3/COEFFICIENT OF X^4 = O
  PRODUCT OF ROOTS = (-1)^4 COEFFICIENT OF X^0/ COEFFICIENT OF X^4 = 54/2 = 27
• 10 years agoHelpfull: Yes(19) No(4)
• the equation can be wriiten as 2*x^4+0*x^3-49*x^2+0*x+54 which is genralised as a*x^4+b*x^3+c*x^2+d*x+e=0
  
  so sum=-b/a=0/2=0 and product=e/a=54/2=27
  
  
• 10 years agoHelpfull: Yes(8) No(0)
• sum of root is: -(-49)/2
  product of root is: 54/2
• 10 years agoHelpfull: Yes(4) No(8)
• sum of the roots = -b/a = 0/2 = 0
  product of roots = d/a = 54/2 = 27
• 10 years agoHelpfull: Yes(2) No(1)
• SUM OF ROOTS = (-1)^4 COEFFICIENT OF X^0/ COEFFICIENT OF X^4 OF X^2/ COEFFICIENT OF X^4= 2-49+54=56-49=7
  PRODUCT OF ROOTS = (-1)^0 COEFFICIENT OF X^3/COEFFICIENT OF X^4 = O
• 10 years agoHelpfull: Yes(0) No(2)
• In General:
  
  Adding the roots gives -b/a (in this case = 54/2 = 27)
  Multiplying the roots gives (where "z" is the constant at the end):
  z/a (for even degree polynomials like quadratics)( in this case = 0)
  -z/a (for odd degree polynomials like cubics)
• 10 years agoHelpfull: Yes(0) No(1)
• 27,0 don't know
• 8 years agoHelpfull: Yes(0) No(0)
• i dont know
  
• 8 years agoHelpfull: Yes(0) No(0)