If the equation (k^2 - 5k + 6)x^2 + (k^2 - 7k + 12)x + k^2 - 9 = 0 satisfy more than two values of x, then the value of k is
option
a) -3
b) 3
c) 2
d) 4

• as more two values of x
  dn we can wrt it (k2-5k+6=0)--> k=3,2
  (k2-7k+12=0) --> k=4,3
  (k2-9=0)--> k=+3,-3
  the common value of k is 3
  and for this value(3) cn get more dn two value of x.
  ans is 3
• 11 years agoHelpfull: Yes(31) No(2)
• Ans is 3.
  [(k-3)(k-2)]x^2+[(k-3)(k-4)]x+k^2-9=0
  The possible values for k are 3,-3,2 & 4.
  the value of k as 3 only satisfies the equation for more than 2 values of x.
• 11 years agoHelpfull: Yes(8) No(1)
• I cant capable to understand what does more than two values depict as it is qudratic so only 2,1 or 0 roots possible.
• 11 years agoHelpfull: Yes(3) No(1)
• roots for (k^2-5k+6): k=2& k=3
  roots for (k^2-7k+12): k=4& k=3
  so the common value is k=3
  ans:3
• 11 years agoHelpfull: Yes(2) No(0)
• Ans is 3.
  [(k-3)(k-2)]x^2+[(k-3)(k-4)]x+k^2-9=0
  The possible values for k are 3,-3,2 & 4.
  the value of k as 3 only satisfies the equation for more than 2 values of x.
• 11 years agoHelpfull: Yes(1) No(0)
• Simply common factor in both equation 3 hence 3 is the correct ans😃
• 11 years agoHelpfull: Yes(1) No(0)
• answer is b)3
• 11 years agoHelpfull: Yes(0) No(0)
• value of k=3 will satisfied.
  bcz of
  (3^2-5+3+6)*x^2+(3^2-7*3+12)*x+3^2-9=0
  (9-15+6)*x^2+(9-21+12)*x+9-9=0
  (15-15)*x^2+(21-21)*x+0=0
  0*x^2+0*x+0=0
  so that answer is k=3
• 11 years agoHelpfull: Yes(0) No(0)
• by option verifiction 3 is the answer
• 11 years agoHelpfull: Yes(0) No(0)