Let f( x ) = ax^2 + bx + c, where a , b and c are certain constants and (a can't be 0). It is known that f(5) = -3f(2) and that 3 is a root of f(x) = 0.
1. What is the other root of f(x) = 0?
(1) -7 (2)- 4 (3) 2 (4) 6 (5) cannot be determined


2. What is the value of a + b + c ?
(1) 9 (2) 14 (3) 13 (4) 37 (5) cannot be determined

• by f(5)=-3f(2)
  37a+11b+4c=0
  by f(3)
  9a+3b+c=0
  multiply second equ. with 4 and subtract it from first
  we will get a=b
  now let z bt the other root of the equ.
  therefore (x-3)(x-z)=ax^2+bx+c
  x^2-(z+3)x+3z=ax^2+bx+c
  a=1
  c=3z
  a=b=1
  b=-(z+3)
  1=-(z+3)
  z=-4 (solution of first part)
  
  now we have an equ by (x-3)(x+4)
  x^2+x-12
  here a=1 is a case
  actual equation is ax^2+bx+c
  where b=a
  c=-12a
  both prooved in first part
  therefore equation is
  ax^2+ax-12a=0
  so a+b+c=a+a-12a=-10a
  therefore cannot be determined
  
• 11 years agoHelpfull: Yes(4) No(0)
• by the f(5)=-3f(2);
  37a+11b+4c=0;
  and
  by f(3)=9a+3b+c=0
  so by multiplying eq2 by 4 andsubtracting from eq 1 then a=b;
  and by the add and multi. of root we can solve other root is -4
• 11 years agoHelpfull: Yes(2) No(1)
• this cant be solved because here we can get only two equations that are
  9a+3b+c=0 ( 3 is a root)and
  25a+5b+c= -3(4a+2b+c)
  37a+11b+4c=0
  we need to have three equation to solve this
  hance it can not be solved
  
• 11 years agoHelpfull: Yes(1) No(2)
• f(5) = −3f(2)
  => 25a+5b+c = -3(4a+2b+c) => 37a+11b+4c = 0 ---(1)
  3 is a root of f(x)
  => f(3) = 0 => 9a+3b+c = 0 ---(2)
  On solving (1) and (2) we get, b = a, c = -12a ---(3)
  Substituting (3) in f(x) = 0 => ax2 + ax – 12a = 0 => a(x2 + x – 12) = 0
  =>(x-3)(x+4) = 0 => x = 3 (or) -4
  
  With conditions (1) and (2), it is not possible to find the value of a+b+c
• 10 years agoHelpfull: Yes(1) No(0)