Quadratic Equation


Direction: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer
( 1 ) if x > y
( 2 ) if x ≥ y
( 3 ) if x < y
( 4 ) if x ≤ y
( 5 ) if x = y or the relationship cannot be established.

  1. Ⅰ. 5x + 2y = 31
    Ⅱ. 3x + 7y = 36











  1. View Hint View Answer Discuss in Forum

    Given that :- Ⅰ. 5x + 2y = 31 .....( 1 )
    Ⅱ. 3x + 7y = 36 .....( 2 )

    Correct Option: A

    Given that :- Given that :- Ⅰ. 5x + 2y = 31 .....( 1 )
    Ⅱ. 3x + 7y = 36 .....( 2 )
    Solving these two linear equations, we get x = 5, y = 3 .
    x > y is correct answer .


  1. Ⅰ. 2x2 + 11x + 14 = 0
    Ⅱ. 4y2 + 12y + 9 = 0











  1. View Hint View Answer Discuss in Forum

    From the given equations , we have
    Ⅰ. 2x2 + 11x + 14 = 0
    ⇒ 2x2 + 7x + 4x + 14 = 0
    ⇒ x( 2x + 7 ) + 2( 2x + 7 ) = 0

    Ⅱ. 4y2 + 12y + 9 = 0
    ⇒ ( 2y + 3 )2 = 0

    Correct Option: C

    From the given equations , we have
    Ⅰ. 2x2 + 11x + 14 = 0
    ⇒ 2x2 + 7x + 4x + 14 = 0
    ⇒ x( 2x + 7 ) + 2( 2x + 7 ) = 0
    ⇒ ( x + 2 )( 2x + 7 ) = 0

    ⇒ x =
    -7
    or -2
    2

    Ⅱ. 4y2 + 12y + 9 = 0
    ⇒ ( 2y + 3 )2 = 0
    ⇒ ( 2y + 3 )( 2y + 3 ) = 0
    ⇒ y =
    -3
    ,
    -3
    2
    2
    From above both equations it is clear that x < y is correct answer .



  1. Ⅰ. x2 - 7x + 12 = 0
    Ⅱ. y2 - 12y + 32 = 0











  1. View Hint View Answer Discuss in Forum

    As per the given above equations , we have
    From equation Ⅰ. x2 - 7x + 12 = 0
    ⇒ x2 - 4x - 3x + 12 = 0

    Ⅱ. y2 - 12y + 32 = 0
    ⇒ y2 - 8y - 4y + 32 = 0

    Correct Option: D

    As per the given above equations , we have
    From equation Ⅰ. x2 - 7x + 12 = 0
    ⇒ x2 - 4x - 3x + 12 = 0
    ⇒ x( x - 4) - 3( x - 4) = 0
    ⇒ ( x - 4) ( x - 3) = 0
    ⇒ x = 4 or, 3
    Ⅱ. y2 - 12y + 32 = 0
    ⇒ y2 - 8y - 4y + 32 = 0
    ⇒ y( y - 8) - 4( y - 8) = 0
    ⇒ ( y - 8) ( y - 4) = 0
    ⇒ y = 4 or, 8
    Thus , x ≤ y is required answer .


  1. Ⅰ. x2 = 729
    Ⅱ. y = √729











  1. View Hint View Answer Discuss in Forum

    According to question ,we have
    From equation Ⅰ.
    x2 = 729
    ⇒ x = ± √729 = ± 27

    From equation Ⅱ.
    y = √729

    Correct Option: D

    According to question ,we have
    From equation Ⅰ. x2 = 729
    ⇒ x = ± √729 = ± 27
    From equation Ⅱ. y = √729
    ⇒ y = 27
    Hence , x ≤ y is required answer .



  1. The equation x2 − px + q = 0, p, q ∈ R has on real root if :









  1. View Hint View Answer Discuss in Forum

    According to question , we can say that
    The equation x2 − px + q = 0, p, q ∈ R has no real root if B2 < 4AC

    Correct Option: B

    According to question , we can say that
    The equation x2 − px + q = 0, p, q ∈ R has no real root
    On comparing with quadratic eq. Ax2 + Bx + C = 0 , we get
    ∴ A =1, B= - p, C = q
    Now ,we have B2 < 4AC
    ⇒ ( - P )2 < 4 x 1 x q
    ⇒ p2 < 4q.
    Hence , option B is correct answer .