If a & b are the root of the equation 4x^2+3x+7=0, then 1/a+1/b=

a)-3/7
b)3/7
c)-3/5
d)3/5
e)-3/4
f)3/4

• answer is a)-3/7
  we know that a+b=-3/4 and ab=7/4
  so 1/a+1/b=(a+b)/ab
  =(-3/7)
• 12 years agoHelpfull: Yes(2) No(0)
• replace x-->1/x
  the equation become 4(1/x)^2+3(1/x)+7=0
  ==> 7x^2+3x+4=0 have two root 1/a & 1/b.
  Sum of the roots= 1/a+1/b = -3/7
• 12 years agoHelpfull: Yes(2) No(1)
• 1/a+1/b =(a+b)/ab
  by quad. eqn formula
  =(-b/a)/(c/a)
  =-3/4 * 4/7=-3/7 ans
• 12 years agoHelpfull: Yes(2) No(0)
• ans:e
  (1/a)+(1/b)=(a+b)/ab.....(1)
  from given eqn
  sum of root=a+b=-3/4
  product of root=7/4
  so,on putting the value in eqn(1)
  ans will be -3/4
  
• 12 years agoHelpfull: Yes(0) No(1)