Quadratic Equation

  1. Which of the following equations is a quadratic ?
    1. x3 - x2 - x + 5 = 0
    2. x4 - 10
    3. 7x2 = 49
    4. x4 - x3 = 9000
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    Quadratic equation must be in the form of ax2 + bx + c = 0, where a ≠ 0.

    Correct Option: C

    Clearly, 7x2 = 49 or 7x2 - 49 = 0, which is of the form ax2 + bx + c = 0, where b = 0.
    Thus, 7x2 - 49 = 0 is a quadratic equation.

    Report

  1. Which of the following equations has real roots ?
    1. 2x2 - 3x + 4 = 0
    2. (x - 1) (2x - 5) = 0
    3. 3x2 + 4x + 5 = 0
    4. Cannot be determined
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    (x - 1) (2x - 5) = 0 ⇒ x = 1, 5/2
    So, its roots are real.

    Correct Option: B

    (x - 1) (2x - 5) = 0 ⇒ x = 1, 5/2
    So, its roots are real.

    Report

  1. Find the roots of the equation 2x2 - 9x - 18 = 0.
    1. 3/2 and 6
    2. -3/2 and - 6
    3. -3/2 and 6
    4. 3/2 and -6
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    Given equation is 2x2 - 9x - 18 = 0
    [by factorisation method]
    ⇒ 2x2 - 12x + 3x - 18 = 0

    Correct Option: C

    Given equation is 2x2 - 9x - 18 = 0
    [by factorisation method]
    ⇒ 2x2 - 12x + 3x - 18 = 0
    ⇒ 2x(x - 6) + 3(x - 6) = 0
    ⇒ (2x + 3) (x - 6) = 0
    ∴ x = -3/2, 6

    Report

  1. Find the roots of the equation 2x2 - 11x + 15 = 0
    1. 3 and 5/2
    2. -3 and -5/2
    3. 5 and 3/2
    4. -5 and -3/2
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    2x2 -11x + 15 = 0
    [by factorisation method]
    ⇒ 2x2 - (6x + 5x) + 15 = 0

    Correct Option: A

    2x2 -11x + 15 = 0
    [by factorisation method]
    ⇒ 2x2 - (6x + 5x) + 15 = 0
    ⇒ 2x2 - 6x - 5x + 15 = 0
    ⇒ 2x(x - 3) - 5 (x - 3) = 0
    ⇒ (2x - 5) (x - 3) = 0
    ∴ x = 5/2, 3
    Hence, the roots are 5/2 and 3.

    Report

  1. If [a + (1/a)]2 = 3 , what is the value of a3 + (1/a)3 ?
    1. 10√3/3
    2. 0
    3. 3√3
    4. 6√3
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    [a + (1/a)]2 = 3
    Taking square roots both sides, we get
    a + (1/a) = √3
    On cubing both sides, we
    [a + (1/a)]3 = (√3)3

    Correct Option: B

    [a + (1/a)]2 = 3
    Taking square roots both sides, we get
    a + (1/a) = √3
    On cubing both sides, we
    [a + (1/a)]3 = (√3)3
    ⇒ a3 + 1/a3 + 3.a.1/[a(a + 1/a)] = 3 √3
    ⇒ a3 + 1/a3 + 3√3 = 3√3
    ∴ a3 + 1/a3 = 0

    Report