I. 3x2-14x+15=0

    II. 15y2-34y+15=0

  • If x ≥ y
  • If x ≤ y
  • If x > y
  • If x < y
  • If relationship between x and y cannot be establishment
  • Explanation:

    I. 3x2-14x+15=0

    3x2-9x-5x+15=0

    3x(x-3)-5(x-3)=0

    (3x-5)(x-3)=0

    ∴ x=5/3,3

    II. 15y2-34y+15=0

    15y2-25y-9y+15=0

    5y(3y-5)-3(3y-5)=0

    (5y-3)(3y-5)=0

    ∴ y = 3/5,5/3

    Hence, x ≥ y

I. 3x2-14x+15=0 II. 15y2-34y+15=0
Home Ask Questions Study Current Affairs Previous Papers Kerala PSCIBPSUPSCRBITNPSCMPSCSSCCBSEUnited StatesModel Tests News More Answers Coaching Centres Careers Downloads Colleges