For what value of “k” will the equation (2kx2 + 5kx +2)=0 have equal roots?
A. 2/7 B. 9/4 C. 16/25 D. 7/1

• for a 2nd degree equation ax2+bx_c=0 has equal roots the condition is b2-4ac =0
  in the given equation
  (5k)^2-4*2k*2=0
  by solving this equation
  we get k=0, k=16/25
• 9 years agoHelpfull: Yes(43) No(0)
• for the roots to be equal b^2-4ac=o is the condition
  
  a=2k b=5k c=2
  
  => (5k)^2-4(2k)(2)=0
  => 25k^2-16k=0
  =>k(25k-16)=0
  
  
  k=16/25;
• 9 years agoHelpfull: Yes(12) No(0)
• for real and equal roots,
  b^2-4ac=0
  k=16/25
• 9 years agoHelpfull: Yes(2) No(3)
• B^2=4AC
  25k^2=4.2K.2
  K=16/25
• 9 years agoHelpfull: Yes(2) No(0)
• roots r equal so b2=4ac
  b=2*root(ac)
  5k=2*root(4k)
  25k2=4*4k
  k=16/25
• 9 years agoHelpfull: Yes(1) No(1)
• for equal roots equation should satisfy b2-4ac=0
  25k2-16k=0
  k=0,k=16/25
  so ans is k=16/25
• 7 years agoHelpfull: Yes(1) No(0)
• option c
  
  b^2=4*a*c for equal roots
  25k^2=16K
  k=16/25
• 8 years agoHelpfull: Yes(0) No(0)
• d2-4ac when root are equal
  so 25k2=16k
  =>k=16/25
• 8 years agoHelpfull: Yes(0) No(0)
• if roots are equal then b^2=4*a*c
  then k=16/25
• 7 years agoHelpfull: Yes(0) No(1)
• 25k^2=16k
  k=0 it is not possible or
  k=16/25
  so Option C
• 7 years agoHelpfull: Yes(0) No(0)
• 16/25
• 6 years agoHelpfull: Yes(0) No(0)