I. x2+28x+192=0

    II y2+16y+48=0

  • If x>y
  • If x ≥y
  • If x
  • If x≤y
  • If x=y or the relationship cannot be established
  • Explanation:

    I. x2+28x+192=0

    ⇒ x2+16x+12x+192 = 0

    ⇒ x(x+16) + 12(x+16)=0

    ⇒ (x+12)(x+16) = 0

    ⇒ x=-12 or -16

    II y2+16y+48=0

    ⇒ y2+12y+4y+48=0

    ⇒y(y+12) + 4 (y+12_ = 0

    ⇒ (y+12)(y+4) =0 ⇒ y=-12 or -4

    Hence, x≤y


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I. x2+28x+192=0 II y2+16y+48=0
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